Just note that $P(Q(x))$ evaluates as \[ P(Q(x)) = P(x^2+cx+d) = (x^2+cx+d)^2 + a(x^2+cx+d)+b = x^4 + 2cx^3 + \text{LOT}, \]where LOT stands for lower order terms. So, $P(Q(x)) - Q(P(x)) = 2(c-a)x^3 + {\rm LOT}$. If $c\ne a$ then $P(Q(x)) - Q(P(x))$ is a degree-3 univariate polynomial, as such, it must have a real root. Thus, $a=c$. Now if $b=d$ as well, then $P(x)=Q(x)$, so any real satisfies $P(Q(x)) = Q(P(x))$. Thus, $b\ne d$.