Is the converse of Cayley-Hamilton Theorem true?
No, the converse of Cayley-Hamilton is not true for $n\times n$ matrices with $n\gt 1$; in particular, it fails for $2\times 2$ matrices.
For a simple counterexample, notice that if $p(A)=0$, then for every multiple $q(x)$ of $p(x)$ you also have $q(A)=0$; so you would want to amend the converse to say "if $p(A)=0$, then $p(a)$ is a multiple of the characteristic polynomial of $A$". But even that amended version is false
However, the only failure in the $2\times 2$ matrix case are the scalar multiples of the identity. If $A=\lambda I$, then $p(x)=x-\lambda$ satisfies $p(A)=0$, but the characteristic polynomial is $(x-\lambda)^2$, not $p(x)$.
For bigger matrices, there are other situations where even this weakened converse fails.
The concept that captures the "converse" of Cayley-Hamilton is the minimal polynommial of the matrix, which is the monic polynomial $p(x)$ of smallest degree such that $p(A)=0$. It is then easy to show (using the division algorithm) that if $q(x)$ is any polynomial for which $q(A)=0$, then $p(x)|q(x)$. (Be careful to justify that if $m(x)=r(x)s(x)$, then $m(A)=r(A)s(A)$; this is not immediate because matrix multiplication is not in general commutative!) So we have:
Theorem. Let $A$ be an $n\times n$ matrix over $\mathbf{F}$, and let $\mu(x)$ be the minimal polynomial of $A$. If $p(x)\in \mathbf{F}[x]$ is any polynomial such that $p(A)=0$, then $\mu(x)$ divides $p(x)$.
The Cayley-Hamilton Theorem shows that the characteristic polynomial is always a multiple of the minimal polynomial. In fact, one can prove that every irreducible factor of the characteristic polynomial must divide the minimal polynomial. Thus, for a $2\times 2$ matrix, if the characteristic polynomial splits and has distinct roots, then the characteristic and minimal polynomial are equal. If the characteristic polynomial is irreducible quadratic and we are working over $\mathbb{R}$, then again the minimal and characteristic polynomials are equal. But if the characteristic polynomial is of the form $(x-a)^2$, then the minimal polynomial is either $(x-a)$ (when the matrix equals $aI$), or $(x-a)^2$ (when the matrix is not diagonalizable).
As for solving this problem: if $\lambda$ is an eigenvalue of $A$, and $A$ is invertible, then $\lambda\neq 0$, and $\frac{1}{\lambda}$ is an eigenvalue of $A^{-1}$: for if $\mathbf{x}\neq\mathbf{0}$ is such that $A\mathbf{x}=\lambda\mathbf{x}$, then multiplying both sides by $A^{-1}$ we get $\mathbf{x} = A^{-1}(\lambda \mathbf{x}) = \lambda A^{-1}\mathbf{x}$. Dividing through by $\lambda$ shows $\mathbf{x}$ is an eigenvector of $A^{-1}$ corresponding to $\frac{1}{\lambda}$.
Since $A=A^{-1}$, that means that if $\lambda_1,\lambda_2$ are the eigenvalues of $A$, then $\lambda_1 = \frac{1}{\lambda_1}$ and $\lambda_2=\frac{1}{\lambda_1}$; thus, each eigenvalue is either $1$ or $-1$.
If the matrix is diagonalizable, then we cannot have both equal to $1$ (since then $A=I$), and they cannot both be equal to $-1$ (since $A\neq -I$), so one eigenvalue is $1$ and the other is $-1$. Since the trace of a square matrix equals the sum of its eigenvalues, the sum of the eigenvalues is $0$.
Why is $A$ diagonalizable? If it has two distinct eigenvalues, $1$ and $-1$, then there is nothing to do; we know it is diagonalizable. If it has a repeated eigenvalue, say $1$, but $A-I$ is not the zero matrix, pick $\mathbf{x}\in \mathbb{R}^2$ such that $A\mathbf{x}\neq \mathbf{x}$; then $$\mathbf{0}=(A-I)^2\mathbf{x} = (A^2-2A + I)\mathbf{x} = (2I-2A)\mathbf{x}$$ by the Cayley Hamilton Theorem. But that means that $2(A-I)\mathbf{x}=\mathbf{0}$, contradicting our choice of $\mathbf{x}$. Thus, $A-I=0$, so $A=I$ and $A$ is diagonalizable. A similar argument shows that if $-1$ is the only eigenvalue, then $A+I=0$. . (Hiding behind that paragraph is the fact that if the minimal polynomial is squarefree and splits, then the matrix is diagonalizable; since $p(x)=x^2-1=(x-1)(x+1)$ is a multiple of the minimal polynomial, the matrix must be diagonalizable).
So this completes the proof that the trace must be $0$, given that $A\neq I$ and $A\neq -I$.
If $A^2 = 1$ then the eigenvalues of $A$ satisfy $\lambda^2 = 1$, so they are either $+1$ or $-1$. As they cannot both be $+1$ or $-1$, we must have one each, and their sum (the trace) is $0$.
If $p(A) = 0$ then $p(A)$ is divisible by the minimal polynomial of $A$. As an extreme example, take $A=0$. Then $p(A) = 0$ for lots of polynomials, but the characteristic polynomial is $x \mapsto x^{\dim A}$.
The multiple choice way to answer this is to note that $\begin{bmatrix}1&0\\0&-1\end{bmatrix}$ is an example with trace $0$, so if the question is valid then the answer must be C.
If $p$ is a polynomial such that $p(A)=0$, then it is true that every eigenvalue of $A$ is a zero of $p$. For if $\lambda$ is an eigenvalue of $A$ with eigenvector $v$, then $0=p(A)v=p(\lambda)v$, which implies $p(\lambda)=0$. From this, you know as Yuval already pointed out that the possible eigenvalues are $1$ and $-1$.
The possible characteristic polynomials are thus $x^2-1$, $(x-1)^2$, and $(x+1)^2$. To rule out the last two cases, you can consider the triangular forms of $A$. For example, having characteristic polynomial $(x-1)^2$ implies that $A$ is similar to a matrix of the form $\begin{bmatrix}1&a\\0&1\end{bmatrix}$. But then the only way for $A=A^{-1}$ to be true would be if $a=0$, contradicting the hypothesis that $A\neq I$.