Finitely many $k$ for which $(1+ap)^k(1+bp)\equiv 1\pmod{p^k}$.

Question

Let $p$ be a prime number, $a,b\in\mathbb Z$ and $ab\ne0$. I guess there exists only finitely many $k\in\mathbb N$ for which $(1+ap)^k(1+bp)\equiv 1\pmod{p^k}$. I think we may do as $(1+bp)((1+ap)^k-1)+bp\equiv 0\pmod{p^k}$. We need to consider the case $p=2$ và $p>2$. I even got stuck in case $p=2$. Any idea?