Let $E=(PQR) \cap MN$, $F=AP\cap BD$ and $G=RE\cap QN$. Note that $M, F, R$ are collinear, indeed $F$ is the orthocenter of $\triangle AMD$ so $MF \perp AD$, but also $ABDC$ is a harmonic quadrilateral so $(A, D; R, Q)=-1$ and since $\angle ABD =90^{\circ}$ we get that BD bisects $\angle PBR$ also notice that PA bisects $\angle BPR$ so $F$ is also incenter of $\triangle PBR$. Now $\angle FPD=90^{\circ}$ so $D$ is actually the P-excenter of $\triangle PBR$. And $\angle FRD=90^{\circ}$ and the claim follows. Easy angle chase leads to $APQN$ cyclic, and $\angle AQN=90^{\circ}$ wich is $MR \| QN$, notice also that since $(A, D; R, Q)=-1$ and $\angle APD =90^{\circ}$ $PD$ bisects $\angle QPR$ and $ER=EQ$ but $\angle AQN=90^{\circ}$ so $ER=EQ=EG$ from here it follows that $MRNG$ is a parallelogram and the conclusion follows

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See my solution to this problem on my Youtube channel here: https://www.youtube.com/watch?v=-Ioqd8PghGI

I thought P1 would be easy, turns out to be not so much. Let $(PQR)$ meet $(ABC)$ at $T$ and $AT$ cut $(PQR)$ at $L$ and $K=AP\cap DT$. 1) By Reim $QL \parallel AB$ and $RL \parallel AC$. And so we have $RQL \sim CBD$ and $L \in PD \in MN$. 2) $D$ - orthocenter of $AKL$, so $KL \perp AD$ and $KL$ - diameter of $(PQR)$. 3) By Pascal on $BPDDAC$ we have $QN \parallel BC$. The same $MR \parallel BC$. In the sum total we have $MR \parallel KL \parallel QN$ and $KL$ bisects $RQ$. By Thales we are done.

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more solutions here: 1st SAFEST Olympiad 2019 p1