It is well-known that $P-$simson line bisects $PH$. Therefore, $R$ is midpoint of $PH$. Let $MN$ further meets $AD$ at $L$. Since $HR=RP$, this implies $MR=RL$. So, $R$ is the midpoint of hypotenuse of right triangle $MDL$. Therefore, $RM=RD=RL$ and $\bigtriangleup DMR$ is isosceles. $\blacksquare$

Let $X=HD\cap MN$ Claim: $\triangle PMR \cong \triangle HXR$ Proof: AA Similarity. Clearly $\angle MRP = \angle XRH$. Next, note that \[\angle RMP = 90 + \angle XMD = 90 + (90-\angle MXD) = \angle RXH\]$\square$. To finish, note that R is the circumcenter of the circle with diameter $MX$, which contains $D$. Thus, $RM=RD$ so $\angle DMR = \angle MDR$ and we're done. $\blacksquare$.

Very easy. It's well known that $R$ is midpoint of $PH$ and let $MN \cap AD=K$ from $PM \parallel AD$ $R$ is center of $MDK$