In the usual context of nonempty structures, the statement is valid, meaning that it is true in all models. One way to see this is that if there is any $y$ in the structure for which $Q(y)$ holds, then the statement is true, since we may always choose that $y$, and this will make $Q(y)$ true and hence the final implication true, regardless of any $x$, and so the whole implication is true. Otherwise, we are in the case where $Q(y)$ is always false, in which case $P(x)\to Q(y)$ is logically equivalent to $\neg P(x)$, and so the quantification over $y$ becomes irrelevent (provided the structure is nonempty). So again the statement is true. So the statement is true in any nonempty structure.
Edit. Meanwhile, in the context of first-order logic allowing the empty structure (which is a bit unusual, and which is only possible if your language has no constant symbols), then the statement is not valid, since it is false in the empty structure. This is because all universal statements hold vacuously in the empty structure and all existential statements fail in the empty structure, so the antecedent is true and the conclusion is false, and so the implication fails in the empty structure.
Conclusion: your statement is valid if the language has constant symbols; valid if your logic disallows the empty structure; but otherwise it is not valid.