Let $\Gamma$ be a circle and $P$ be a point outside, $PA$ and $PB$ be tangents to $\Gamma$ , $A, B \in \Gamma$ . Point $K$ is an arbitrary point on the segment $AB$. The circumscirbed circle of $\vartriangle PKB$ intersects $\Gamma$ for the second time at point $T$, point $P'$ is symmetric to point $P$ wrt point $A$. Prove that $\angle PBT = \angle P'KA$.
Problem
Source: Ukrainian Geometry Olympiad 2020, IX p2, X p1, XI p1
Tags: geometry, equal angles, circles, tangent, symmetry