If $b=0$ then $P(x-a)=P(x)\Leftrightarrow a=0$.So $b\neq 0$ For $x=0$ we have that $P(0)=0$ If $q\neq 0$ a root of $P(x)$ then $P(q-a)=0$ so $q-a$ is also root of $P(x)$ This means that $q=ak,k\in \mathbb{Z}$. If $a>0$ clearly $k>0$.If $a<0$ we must have $|q-a|<|q|$ so $q<0$ and $k>0$. So $k\in \mathbb{N}$.This means that the roots of $P(x)$ are $0,a,2a,..,ka$ For $x=(k+1)a$ we must have that $b=(k+1)a$ So if $b\neq (k+1)a$ for some $k\in \mathbb{N}$ there are no solutions If there exist $k\in \mathbb{N} $ such that $b=(k+1)a$ then $P(x)=mx\prod_{i=1}^{k} (x-ia),m\in \mathbb{R}$