BALTIC-WAY 2025 Problem 14

In an acute triangle $ABC$ with $AB < AC$, the circumcircle is $\omega$ and the orthocentre is $H$. The point $P$ on major arc $BC$ of $\omega$ satisfies $\angle PCB + \angle ACB = 90^\circ$. The point $R$ on segment $AC$ satisfies $BR = CR$. Prove that $\angle HPR = \angle ACB$.