Solution:

The answer is (E). Let $m, n$ be the degrees of $p(x)$ and $q(x)$, respectively. These are two positive integers. The relation between the polynomials given in the problem implies the following equation on the degrees, $m n = 3 m + 5 n$, which in turn is equivalent to $(m-5)(n-3) = 15$. To find $(m, n)$ we must therefore solve the systems
$$\left\{ \begin{array}{l} m-5 = k \\ n-3 = \frac{15}{k} \end{array} \right.$$
as $k$ ranges over the divisors of $15$, that is $k = \pm 1, \pm 3, \pm 5, \pm 15$. Note however that, when $k$ is negative, at least one of $m$ and $n$ turns out to be negative or zero. Therefore we discard these cases and consider only the remaining four systems, from which we obtain the following solutions:

Now we can observe that for each pair $(m, n)$ there actually exist two polynomials $p(x) = x^{m}$, $q(x) = (x-1)^{n}$ of degrees $m, n$ that satisfy the given relation. With these examples we can already exclude answers $(B)$ and $(C)$. As for $(A)$ and $(D)$, it suffices to observe that, if $(p(x), q(x))$ satisfies Marcella's condition, the same holds for the polynomials $(\lambda p(x), q(x))$ for any integer $\lambda \neq 0$: hence $\left(-x^{m}, (x-1)^{n}\right)$ and $\left(2 x^{m}, (x-1)^{n}\right)$ are two easy counterexamples. Finally, the degree of $p(x) q(x)$ is $m+n$ and can only be $16$ or $24$, which establishes the correctness of (E).