US-TST 2024 Problem 2

Let $A B C$ be a triangle with incenter $I$ . Let segment $A I$ intersect the incircle of triangle $A B C$ at point $D$ . Suppose that line $B D$ is perpendicular to line $A C$ . Let $P$ be a point such that $\angle B P A = \angle P A I = 90^{\circ}$ . Point $Q$ lies on segment $B D$ such that the circumcircle of triangle $A B Q$ is tangent to line $B I$ . Point $X$ lies on line $P Q$ such that $\angle I A X = \angle X A C$ . Prove that $\angle A X P = 45^{\circ}$ .