Solution:

The answer is $(\mathbf{C})$. $P_{2}$ can be written as a product of first-degree factors, so it will be of the form
$$\left(x-\alpha_{1}\right)^{m_{1}} \ldots\left(x-\alpha_{k}\right)^{m_{k}}$$
where $\alpha_{1}, \ldots, \alpha_{k}$ are all and only the solutions of the equation $P_{2}(x)=0$. Since $P_{1}$ divides $P_{2}$, it follows that $P_{1}$ can also be written as a product of first-degree factors, and in particular as a product of the same factors as $P_{2}$ but with lower or equal multiplicity, that is
$$P_{1}=\left(x-\alpha_{1}\right)^{n_{1}} \ldots\left(x-\alpha_{k}\right)^{n_{k}}$$
with $n_{1} \leq m_{1}, \ldots, n_{k} \leq m_{k}$. Since $P_{1}$ has degree strictly less than $P_{2}$, at least one of the previous relations must be a strict inequality, which implies that there exists at least one first-degree factor that divides both $P_{2}$ and $Q$, that is, there exists at least one real number $a$ such that $P_{2}(a)=Q(a)=0$.