Solution:

Take $X$ on $AH$ so that $BX$ is perpendicular to $AH$. Extend to meet $HM$ at $P'$. Let $N$ be the midpoint of $AB$. $A$, $B$, $M$ and $X$ are on the circle center $N$ radius $NA$ (because angles $AMB$ and $AXB$ are $90^{\circ}$). Also $MN$ is parallel to $BC$ (because $AMN$, $ACB$ are similar), so $NM$ is perpendicular to $MH$, in other words $HM$ is a tangent to the circle. Hence $P'M = P'X$. $P'B$. Triangles $P'XH$ and $P'HB$ are similar (angles at $P'$ same and both have a right angle), so $P'H / P'X = P'B / P'H$, so $P'H \cdot P'H = P'X \cdot P'B$. Hence $P'H = P'M$ and $P'$ coincides with $P$.