We let $\{Q\} = HS \cap AC$ and $\{I,R\} = IB \cap (HAC)$. Let $s$ be a line perpendicular to $r$ through $H$ which intersects $r$ and $(HAC)$ at $Q'$ and $S' \neq H$, respectively. Since $HR$ is a diameter of $(HAC)$ (because $\angle HIR = 90^{\circ}$), it follows that $\angle HS'R = 90^{\circ}$, so $S'R \parallel AC$ and $AS'RC$ is an isosceles trapezoid. That with $AB=BC$ gives us $BS'=BR$, which in turn gives us $\angle S'BP = \angle Q'BI$. But since $\angle HIB = 90^{\circ} = \angle HQ'B$, $HIQ'B$ is cyclic, so $\angle Q'BI = \angle IHQ' = \angle IHS'$. We can now see that $P,H,I$ being collinear implies $\angle S'BP = \angle IHS' = \angle PHS'$, which would mean $S'$ is in $(HPB)$ and $(HAC)$ so $S'=S$ and $HS \perp r$. And $HS \perp r$ implies that $S'=S$ and $Q'=Q$ which implies $\angle PHS = \angle PBS = \angle QBI = \angle IHS$, which means $P,H,I$ are collinear and we are done.