SHARYGIN 2018 Problem 16

Let $ABC$ be a triangle with $AB < BC$ . The bisector of angle $C$ meets the line parallel to $AC$ and passing through $B$ , at point $P$ . The tangent at $B$ to the circumcircle of $ABC$ meets this bisector at point $R$ . Let $R'$ be the reflection of $R$ with respect to $AB$ . Prove that $\angle R'P B = \angle RPA$ .