Intuition for the infinite cardinals p and t (now that p = t)?
A couple of simple examples; they don't have cardinality $\mathfrak p$ or $\mathfrak t$ but depending on your knowledge they may or may not help.
They are families of sets of natural numbers that I'll call $\mathcal F$, $\mathcal L$, $\mathcal M$, and $\mathcal D$.
Let $\mathcal F=\{X\subseteq\mathbb N: X\text{ is co-finite}\}$ and $\mathcal L=\{[n,\infty)\cap\mathbb N: n\in\mathbb N\}$.
In terms of the definitions on page 2 of
M. Malliaris, S. Shelah, Cofinality spectrum theorems in model theory, set theory and general topology, arXiv:1208.5424 (or direct pdf link)
$\mathcal L$ is not a tower, because even though it is well-ordered by $\supseteq^*$, it has $\mathbb N$ itself as a pseudo-intersection (the elements of $\mathcal L$ are too big, so to speak).
Anyway, $\mathcal L$ is countable, so it's plausible that a tower would have to be uncountable, and indeed that's the cardinality $\mathfrak t$.
$\mathcal F$ is a bit bigger than $\mathcal L$, is not well-ordered by $\supseteq^*$, but still has the s.f.i.p. since the intersection of finitely many co-finite sets is still co-finite (since the union of finitely many finite sets is finite).
Like $\mathcal L$, the family $\mathcal F$ is countable and unfortunately has $\mathbb N$ as pseudo-intersection. So it's again plausible that the corresponding cardinal $\mathfrak p$ is uncountable.
So let's consider $\mathcal M$, the family of sets whose limiting (Banach) density is 1. It has the s.f.i.p. and no pseudo-intersection. Great! However, $\mathcal M$ has cardinality $2^{\aleph_0}$.
Now consider $\mathcal D=\{\{k: k\text{ is divisible by }2^m\}: m\in\mathbb N\}$. This is also countable, well-ordered by $\supseteq$, but again unfortunately has an infinite pseudo-intersection, namely $\{1,2,4,8,16,\dots\}$.
$\begin{eqnarray*} \text{Family }\quad & \text{Well-ordered?}\quad&\text{S.f.i.p.? }\quad & \text{pseudo-$\cap$? }&\text{Cardinality}\\ \mathcal F & \text{no}& \text{yes} &\text{yes}& \aleph_0\\ \mathcal L & \text{yes}& \text{yes} &\text{yes}& \aleph_0\\ \mathcal M & \text{no}& \text{yes} &\text{no}& 2^{\aleph_0}\\ \mathcal D & \text{yes}& \text{yes} &\text{yes}& \aleph_0\\ \end{eqnarray*}$
Each element of p and each element of t is a countable family of infinite sets of natural numbers. Let's start by giving an example of such family.
A simple example of a countable family of infinite sets of natural numbers would be $$E = \bigotimes_{j\in \Bbb N} \{k \in \Bbb N : \exists m | k = jm \}$$ so that two elements of $E$ would be $\{ 2,4,6,8\cdots\}$ and $\{ 3,6,9,12\cdots\}$. We can say that a countable family of infinite sets of natural numbers is a subset of $[\Bbb N]^{\aleph_0}$. (We will see that this particular $E$ is not in p.)
Now let's define a relationship between sets $A$ and $B$ which I will call "is almost a subset of," written as $A \subseteq^* B$. In words, $A$ is almost a subset of $B$ iff only a finite set of elements of $A$ are not also in $B$. That is, $$A \subseteq^* B \Leftrightarrow |\{ x:x\in A \wedge x \not\in B\}| < \infty $$ For example if $A$ is the set of primes and $B$ is the set of odd natural numbers, then $\{ x:x\in A \wedge x \not\in B\} = \{2\}$ and $A \subseteq^* B$.
Note that it is quite possible for $A \subseteq^* B$ and $B \subseteq^* A$.
Now let's define p and give an example of one element of p.
p is the set of all countable families of infinite sets of natural numbers such that family $f \in \mathbf{p}$ if and only if two conditions are met:
Every non-empty finite subfamily of $f$ (every finite collection of sets, all of which are in $f$) has an infinite intersection.
But there is no infinite set of integers $A$ such that $A$ is almost a subset of each element of $f$. That is, $\left( \forall A \subseteq \Bbb N: B \in f \implies A \subseteq^* B \right) \implies |A| = \infty$.
You can see that these two conditions work against one another in some sense. A simple example of a family of sets $f_1$ meeting the first condition is $\left\{ \{ n+0 :n \in \Bbb N\},\{ n+1 :n \in \Bbb N\}, \cdots \right\}$ since the intersection of any finite number of elements of $f_1$ consists of all numbers $\geq$ the highest starting point of any of that finite set of elements.
But that family $f_1$ fails the second condition, because if $A$ is the (infinite) set of even integers, then $A$ is almost a subset of each member of $f_1$.
An example of an element of p is the family of sets (indexed by $k\in \Bbb N$) of the form $\{ m^k : k\in \Bbb N\}$. The first condition is met since if $\ell$ is the least common multiple of all the indices in the finite sub-family then the set $\{ m^\ell : m\in \Bbb N\}$ is in every set in the finite subfamily. The second condition is met since there are an infinite number of integers that are cubes but not squares and vice-versa, so any infinite set $A$ of integers is either not almost a subset of the squares, or not almost a subset of the cubes, unless all but a finite number of elements of $A$ are sixth powers. But then $A$ won't be almost a subset of the fifth powers, unless all but a finite number of elements of $A$ are $30$-th powers; and so forth.
Now let's define t and give an example of one element of t.
t is the set of all countable families of infinite sets of natural numbers that can be well-ordered by the $\subseteq^*$ relation. (Remember, well-ordering will mean that if family $T$ is in t then every (non-equivalent under $A \subseteq^* B$) pair of elements of $T$ can be related by $\subseteq^*$ in one direction or the other, and every subset of $T$ has a least element in that ordering.)
A simple example of an element of t is the family $f_1$ described above. This "tower" happens not to be in p.
The p = t question, then, is whether p can be mapped onto t, or t can be mapped onto p, or they are in $1:1$ correspondence.
In the paper by Malliaris and $(\mbox{Shelah})^2$ there is a minor wording error in definition 1.1 whch might be causing some confusion: The first bullet should read
$D$ has an infinite pseudo-intersection if there is an infinite $A\subseteq \Bbb N$ such that $\forall B\in D, A \subseteq^* B$.