Let $X$ be the point of intersection of the lines $BH$ and $AC$, and let $N$ be the midpoint of the line segment $AH$. Since $\angle AXH = \angle APH = 90^\circ$, the points $P$, $X$ lie on the circle having $AH$ as its diameter (if $AB = AC$, then $P$ coincides with $H$ and it is clear that $X$ lies on the circle with $AH$ as its diameter in this case). Since $N$ is the midpoint of $AH$, we then have $\angle AXN = \angle XAN$. Also, since $\angle BXC = 90^\circ$, the point $X$ lies on the circle having $BC$ as its diameter, and from this we get $\angle CXM = \angle XCM$ and $XM = BM$. Using these facts we get
$$\angle NXM = 180^\circ - (\angle AXN + \angle CXM) = 180^\circ - (\angle XAN + \angle XCM) = 90^\circ,$$
from which it follows that the circle going through the 3 points $A, P, X$ is tangent to the line $MX$. Hence by the well-known theorem on the power of a point with respect to a circle, we obtain $AM \cdot PM = (XM)^2 = (BM)^2$.