Showing equivalent metrics have the same convergent sequences
If you’ve really already proved that metrics $p$ and $d$ related in that way generate the same open sets, you’re practically done, but you’re trying to make it much too complicated. Suppose that $\langle x_n:n\in\Bbb N\rangle$ converges to $x$ with respect to $d$; you want to show that it converges to $x$ with respect to $p$ as well. Let $U$ be an open nbhd of $x$. Then since $\langle x_n:n\in\Bbb N\rangle\underset{d}\longrightarrow x$, there is an $m\in\Bbb N$ such that $x_n\in U$ for all $n\ge m$. But that’s also exactly what it means for $\langle x_n:n\in\Bbb N\rangle$ to converge to $x$ with respect to $p$, so $\langle x_n:n\in\Bbb N\rangle\underset{p}\longrightarrow x$.
It’s in the proof that $d$ and $p$ generate the same topology that you would use the constants $k$ and $t$. But it’s not necessary to prove first that $d$ and $p$ generate the same topology: you can prove this result directly.
Suppose that $\langle x_n:n\in\Bbb N\rangle\underset{d}\longrightarrow x$. Then for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $d(x_n,x)<\epsilon$ for each $n\ge m_\epsilon$. This immediately implies that $p(x_n,x)<t\epsilon$ for each $n\ge m_\epsilon$. Thus, for each $n\ge m_{\epsilon/t}$ we have $p(x_n,x)<t\cdot\frac{\epsilon}t=\epsilon$, and it follows that $\langle x_n:n\in\Bbb N\rangle\underset{p}\longrightarrow x$. The opposite implication is proved similarly, using the fact that $d(x,y)\le\frac1kp(x,y)$ for all $x,y\in X$.
Note that your description $$kd(x,y)\leq p(x,y)\leq td(x,y)\tag{1}$$ means that $p$ and $d$ are strongly equivalent metrics. For such metrics, it's fairly easy to show that we have the same convergent sequences.
Let $(x_n)$ be any sequence of points of $X$.
On the one hand, suppose that $p(x_n,x)\to0$ for some $x\in X$. Take any $\epsilon>0$. To show that $d(x_n,0)\to 0$, we must show that there exists some $N_1$ such that $d(x_n,x)<\epsilon$ for all $n\geq N_1$. Since $p(x_n,x)\to 0$, then there exists $N_1$ such that $p(x_n,x)<k\epsilon$ for all $n\geq N_1$, and so by $(1)$, $$d(x_n,x)=\frac1k\cdot kd(x_n,x)\leq\frac1k\cdot p(x_n,n)<\frac1k\cdot k\epsilon=\epsilon$$ for all $n\geq N_1$.
On the other hand, suppose $d(x_n,x)\to 0$ for some $x\in X$. Taking $\epsilon>0$ and finding $N_2$ such that $d(x_n,x)<\frac{\epsilon}t$ for all $n\geq N_2$, we again use $(1)$ to see that $p(x_n,x)<\epsilon$ for $n\geq N_2$.
Thus, the metrics share the same convergent sequences.
It isn't actually necessary to use the constants $k,t$--in fact, it's even okay if there are no positive constants $k,t$ satisfying $(1)$ for all $x,y\in X$.
We say that metrics are equivalent if they produce the same open sets. Put rigorously, we mean the following:
(i) $\forall x,y,z\in X$ $\forall R>0$, if $d(x,y)<R$, then there exists $r>0$ such that $d(x,z)<R$ whenever $p(y,z)<r$. (Every point in an open $d$-ball lies in an open $p$-ball contained in the $d$-ball.)
(ii) [Same thing as (i), but swapping $p$ and $d$ in each instance.]
Strongly equivalent metrics are equivalent, but the converse needn't hold.
Suppose that $p$ and $d$ are equivalent, and let $(x_n)$ is a sequence of points of $X$.
On the one hand, suppose $p(x,x_n)\to 0$, and take $\epsilon>0$. Since $p$ and $d$ are equivalent, then there is some $r>0$ such that $d(x,x_n)<\epsilon$ whenever $p(x,x_n)<r$. Since $p(x,x_n)\to 0$, there is some $N$ such that $p(x,x_n)<r$ for all $n\geq N$, so $d(x,x_n)<\epsilon$ for all $n\geq N$. Thus, $d(x,x_n)\to 0$.
A similar approach will let you show that $d(x,x_n)\to0$ implies $p(x,x_n)\to 0$.
Keep in mind that in a metric space with metric $d$, a sequence $(x_n)_{n=1}^{\infty}$ converges to $x \Leftrightarrow$ if $U$ is an open set (with respect to the topology generated by $d$) containing $x$ then $\exists N \in \mathbb{Z}^+$ such that $x_n \in U$ for all $n \geq N$.
So, suppose $(x_n)_{n=1}^{\infty}$ converges to $x$ with respect to the metric $d$. Let $U$ be an open set (in the topology induced by $p$) containing $x$. $U$ is open in the topology generated by $d$. (You already know that equivalent metrics generate the same topology.) Thus $\exists N \in \mathbb{Z}^+$ such that $x_n \in U$ for all $n \geq N$. This shows that $(x_n)_{n=1}^{\infty}$ converges to $x$ with respect to the metric $p$.
You can also go in the other direction, to show that convergence to $x$ with respect to $p$ implies convergence with respect to $d$.