NL-BXMO-TST 2019 Problem 2

Let $\Delta ABC$ be a triangle with an inscribed circle centered at $I$ . The line perpendicular to $AI$ at $I$ intersects $\odot (ABC)$ at $P,Q$ such that, $P$ lies closer to $B$ than $C$ . Let $\odot (BIP) \cap \odot (CIQ) =S$ . Prove that, $SI$ is the angle bisector of $\angle PSQ$