Nice problem Let $N$ be the midpoint of $DE$ (which coincides with the midpoint of $AO$), $X, Y$ be the projection of $B, C$ on line $DE$. Easy length-chasing with $XP=XR, YP=YR$ leads : $XP=YN$. Hence, $MP=MN$ ; or $P, N$ lies on the circle $(M, MB)$. Consider the radical axis theorem on circles $(O), (ANP), (BCNP)$ imply that $PN, BC$ and the tangent from point $A$ at $(O)$ are concurrent. Say that they concur at $R$. Then $\angle RAO=\angle OMR \implies (R, A, M, O) \text{ are concylic } \implies \angle RMA=\angle ROA$, and also $NO^2=NA^2=NP \times NR \implies \angle RON=\angle NPO$. Hence, $\angle RMA=\angle NPO$, as desired. $\blacksquare$

Extension of Grade 10 P3. Nevertheless, still a beautiful problem. CSEMO 2018 Grade 11 Q3 wrote: Let $O$ be the circumcenter of $\triangle ABC$, where $\angle ABC> 90^{\circ}$ and $M$ is the midpoint of $BC$. Point $P$ lies inside $\triangle ABC$ such that $PB\perp PC$. $D,E$ distinct from $P$ lies on the perpendicular to $AP$ through $P$ such that $BD=BP, CE=CP$. If quadrilateral $ADOE$ is a parallelogram, prove that $$\angle OPE = \angle AMB.$$ Let $F$ be the midpoint of $DE$, $I$ be the intersection of $BD$ and $CE$, and $J$ be the intersection of $BC$ and $DE$. Since $ADOE$ is a parallelogram, $AO$ bisects $DE$. Claim 01. $BPFC$ is cyclic. Proof. Notice that $DB \perp CE$. Since $F$ is the midpoint of $DE$, we have \[ \measuredangle FIB \equiv \measuredangle FID = \measuredangle IDF \equiv \measuredangle BDP = \measuredangle DPB = \measuredangle FPB \]Therefore $BPFI$ is cyclic. Similarly, we get $BFCI$ being cyclic. Since $\measuredangle BIC = \measuredangle BPC = 90^{\circ}$ as well, we have $BPCI$ is cyclic. Therefore, we conclude that $BPFC$ is cyclic. Claim 02. $AJ$ tangent to $(ABC)$. Proof. By radical axis theorem on $(ABC), (APF), (BPFC)$. We have $PF \cap BC = J$ lie on the radical axis of both $(ABC)$ and $(APF)$. However, $(APF)$ has diameter $AF \equiv AO$, and therefore these circles are tangent to each other at $A$. Hence, $AJ$ is the radical axis of these circle, proving that $AJ$ is tangent to $(ABC)$. Claim 03. $OA$ tangent to $(JPO)$. Proof. First, we claim that $OA$ is tangent to $(JAP)$. This is obvious as $(JAP)$ has diameter $AJ$, which is perpendicular to $AO$. Since $DE$ bisects $AO$, $F$ is the midpoint of $AO$. Now, we have \[ FJ \cdot FP = \text{Pow}_F (JAP) = FA^2 = FO^2 \]Therefore, $AO$ is tangent to $(JPO)$. Claim 04. $OAJM$ is cyclic. Proof. This follows from $\measuredangle OAJ = 90^{\circ} = \measuredangle OMJ$. To conclude, notice that \[ \measuredangle OPE \equiv \measuredangle OPF = \measuredangle FOJ = \measuredangle AOJ = \measuredangle AMJ= \measuredangle AMB \]