Troll. Actually $I$ is the orthocenter of $\triangle BRC$. Let the circle with diameter $BC$ meet $BI$ at $L$ and redefine $P=MK \cap CL$. Since $LPMJ$ is cyclic with diameter $JP$, we have: $$\angle AJP=\angle AJI+\angle LJP=\angle PMC+\angle LMP=\angle LMC=\angle ABC, $$so the new definition is correct. Thus, $BI \perp CP$ and similarly $CI \perp BQ$, which finishes the problem. @below, perhaps you're right, but at first I started with adding the incircle touchpoint and using the fact that it lies on $(KMJ)$ and I tried some more complicated things to prove it lies on $IR$, e.g. I noticed that $(AMB) \cap (AJP)\in MK$. When I thought about that $I$ might be the orthocenter and found that it was indeed true, I was slightly disappointed, so that's why I said it's troll.

After suspecting that $I$ is the orthocenter (@above, I've seen such a thing before several times, so wouldn't say it's trolling), the actual problem becomes as follows: Let $ABC$ be a triangle with midpoint $M$ of $BC$ and incenter $J$ of triangle $ABM$. The point $P$ is such that $\angle AJP = \angle ABC$ and $\angle JMP = 90^\circ$. Prove that the lines $BJ$ and $CP$ are perpendicular. The reduced problem follows by combining Sine Law in triangles $BMJ, CMP, JMP$ (note that all angles are computable as expressions of $\angle ABC = \beta$, $\angle BMJ = \varphi$ and $\angle MCP = x$).

Intersecting the lines $AJ$ and $MP$ at $T$ yields that $BJMT$ is cyclic quadrilateral and $BPCT$ is parallelogram. Similarly define $S=AK\cap MQ$ and found that $KCSM$ is cyclic and $BQCS$ is parallelogram. Hence, using $\angle BCS=\angle QBC,$ one can find that $CI\perp BR.$ Similarly, $BI\perp CR.$ Therefore, $I$ is the orthocenter of $BRC.$

We uploaded our solution https://calimath.org/pdf/MEMO2024-T6.pdf on youtube https://youtu.be/vHCURoJoTHU.

Note that $\angle PMJ=90^{\circ}$ Motivated Walkthrough of sorts:- The problem is equivalent to proving $I$ is the orthocenter of $\triangle CBR$. Let $A'$ be the relfection of $A$ over $M$, then $P$ is the $A'-$excenter w.r.t $\triangle A'MC$ iff the reflection of $J$ over $M$ is collinear with $A', P$. Since $\angle B=\angle AJP$ we have $\angle MJP=90^{\circ}-\frac{\angle B}{2}=\angle MJ'P$ and $\angle MJA=90^{\circ}+\frac{\angle B}{2}=\angle MJA'$ so $\overline{A'-J'-P}$ is collinear. So we get $BI$ and $CP$ will be perpendicular to each other and similarly we will also have $CI\perp BQ$ and $I$ is the orthocenter of $\triangle CBR$.

motivation