Solution:

For the sake of simplicity, we can assume the balls and bins are both distinguishable.

Since each case includes having $4$ balls in $3$ bins, we can leave those out during our calculation.

For $p,$ there are $5$ choices for the bin with $3$ balls and then $4$ choices for the bin with $5$ balls. Finally, there are $\binom{8}{3} = 56$ ways to choose which balls go in the bins.

For $q,$ after cancelling out $3$ of the $4$ s, there are $\binom{8}{4} = 70$ ways to ensure $4$ balls go in each of the remaining bins.

Since the total number of distributions is the same for both $p$ and $q,$ we can let $\dfrac{p}{q}$ be the ratio of the numerators. Therefore, $$\dfrac{p}{q} = \dfrac{20 \cdot 56}{70} = 16.$$

Thus, E is the correct answer.