The answer are all primes. First, for $p=2$, $P(x)=Q(x)=x^2$ works. Furthermore, for $p=3$, take $P(x) = Q(x)=x^2+1$ works. Assume $p\ge 5$ in the remainder and write $P(x)=x^2+ax+b$ and $Q(x)=x^2+cx+d$. Then, \[ P\cdot Q (x) = x^4 + (a+c)x^3 + (ac+b+d)x^2 + (ad+bc)x + bd. \]Thus, we demand $c\equiv -a\pmod{p}$, $-a^2+b+d\equiv -16\pmod{p}$, $a(d-b)\equiv 0\pmod{p}$ and $bd\equiv 4\pmod{p}$. Case 1. $(3/p)=1$ or $(5/p)=1$. Suppose $(3/p)=1$. Take $b\equiv d\equiv -2\pmod{4}$. With this, it suffices to ensure $-a^2-4\equiv -16\Leftrightarrow a^2\equiv 12\pmod{p}$ is solvable, which is indeed the case by the assumption $(3/p)=1$. Likewise, the case $(5/p)=1$ is handled by taking $b=d=2$ and $a^2\equiv 20\pmod{p}$, which has a solution. Case 2. $(3/p)=(5/p)=-1$. In this case, $(15/p)=1$. Now we take $a\equiv c\equiv 0\pmod{p}$; it boils down ensuring $b,d$ such that $b+d\equiv -16\pmod{p}$ and $bd\equiv 4\pmod{p}$ exists. Take $b\equiv \frac4d\pmod{p}$ (clearly the inverse of $d$ modulo $p$ exists). Then, it suffices to ensure $d+4/d \equiv -16\pmod{p}$ is solvable, that is $(d+8)^2 - 60\equiv 0\pmod{p}$ is solvable. But $(60/p)=(4/p)(15/p)=(15/p)=1$, so this congruence do have a solution. Case closed.