WLOG $P,Q$ are monic. Since $\deg (P \cdot Q)=4032$ and $P(Q(x))$ has exactly $4032$ real roots, therefore $P(x)$ has exactly $2016$ roots. Let $x_1,x_2,...,x_{2016}$ be the real roots of $P(x)$. Then $P(Q(x))=\displaystyle \prod _{i=1}^{2016} (Q(x)-x_i)$ Because $P(Q(x))=0$ has $4032$ real roots, therefore each equation $Q(x)-x_i=0 (i=\overline{1,2016})$ has $2$ real roots. Let $y_{2i-1},y_{2i}$ be the real roots of $Q(x)-x_i=0$. We have $\lbrace y_1,y_2,...,y_{4032} \rbrace = \lbrace -2015,-2014,\cdots ,2017 \rbrace $ and $y_1+y_2=y_3+y_4=\cdots =y_{4031}+y_{4032}=y$ (since all the equation $Q(x)-x_i=0$ has the same second and leading coefficients) Therefore $2016y=y_1+y_2+\cdots +y_{4032}=2016+2017=4033 \Rightarrow y=\dfrac{4033}{2016}$ But $y_i \in \mathbb{Z} \forall i=\overline{1,4032}$ and $y=y_1+y_2$, hence $y \in \mathbb{Z}$ (contradiction) Therefore, it is not possible.