Solution:

The answer is $\mathbf{(C)}$. Let us call $q_{k}(x) = p(p(\ldots(p(x))))$ the polynomial obtained by composing the polynomial $p$ with itself $k$ times, so that $q(x) = q_{2024}(x)$. It is clear that $q_{k}(x)$ is a polynomial with integer coefficients. Since for $k > 1$ we have $q_{k}(x) = p\left(q_{k-1}(x)\right)$ and since the only solutions to the equation $p(x) = 0$ are given by $x = 0$, $x = -1$ and $x = 2$, we have that $q_{k}(x) = 0$ if and only if $q_{k-1}(x) \in \{0, -1, 2\}$.

We observe that the equation $p(x) = -1$ has no integer solutions, while $p(x) = 0$ (resp. $p(x) = 2$) has as its only solution $x = 0$ (resp. $x = 1$); therefore, for $k \geq 2$ we have $q_{k-1}(x) = p\left(q_{k-2}(x)\right) \in \{0, 1, 2\}$ if and only if $q_{k-2}(x) \in \{0, 1\}$.

Finally, since $p(x) = 1$ has no integer solutions, we have that for $k > 3$ the equation $q_{k-2}(x) = p\left(q_{k-3}(x)\right) = 1$ has no solutions and therefore $q_{k-1}(x) = p\left(q_{k-2}(x)\right) \in \{0, 1, 2\}$ if and only if $q_{k-2}(x) = 0$.

Summing up the above, for $k > 3$, we have that $q_{k}(x) = 0$ holds if and only if $q_{k-2}(x) = 0$. Concatenating this implication enough times, we obtain that $q(x) = q_{2024}(x) = 0$ if and only if $q_{2}(x) = p(p(x)) = 0$, which can easily be verified to have as solutions $x \in \{-1, 0, 1, 2\}$.