Is the complement of this decision problem in $P$?
Careful how you come up with a complement. The original problem asks to satisfy the following, given M:
$$\exists p,q: ((p\text{ and } q \text{ are prime})\land \neg(p\mid M)\land \neg(q\mid M) \land (pq=M))$$
As you say, the answer is always NO.
The complement of the problem should by definition have the opposite answer (YES) for ever $M$. To see why that's the case, negate the original problem (carefully):
$$\neg\exists p,q: ((p\text{ and } q \text{ are prime})\land \neg(p\mid M)\land \neg(q\mid M) \land (pq=M))$$
Distributing through the $\exists$ gives:
$$\forall p,q: \neg((p\text{ and } q \text{ are prime})\land \neg(p\mid M)\land \neg(q\mid M) \land (pq=M))$$
And distributing again makes this pretty trivial:
$$\forall p,q: ((p\text{ or } q \text{ is not prime})\lor (p\mid M)\lor (q\mid M) \lor (pq\ne M))$$
So the complement is really asking: Given $M$, is it true that every pair of numbers $p$ and $q$ either has a non-prime or a divisor of $M$ or they don't multiply to $M$? It's a bit strange to claim this, but it's always true.