Firstly, we know that if $k$ is odd, then for any integer $m$, we must have $km\equiv m\pmod{2}$.

Let $P(x) = a_nx^n + a_{n-1}x^{n-1}+\cdots + a_1 x + a_0$. Then $$P(2015) \equiv a_n + a_{n-1} + \cdots + a_1 + a_0 = P(1) =2015\equiv 1\pmod{2}$$

But $2016$ is even, Therefore it is impossible for $P(2015)=2016$.