Solution:
Note that $MK \perp MJ$. By simple angle chasing we get that
$$\begin{gathered} \angle PJM = \angle AJM - \angle AJP = 90^\circ + \frac{1}{2} \angle ABC - \angle ABC = 90^\circ - \frac{1}{2} \angle ABC, \\ \angle JPM = 90^\circ - \angle PJM = \frac{1}{2} \angle ABC . \end{gathered}$$
Let $P'$ be the reflection of $P$ over $M$. Note that $\angle JPM = \angle JP'M$ as $JM \perp MP$. Then
$$\angle JP'M = \angle JPM = \frac{1}{2} \angle ABC = \angle JBM$$
hence $JBP'M$ is concyclic and as $MK \perp MJ$, point $P'$ is the $A$-excenter of triangle $ABM$ by the incenter-excenter lemma. Let $Q'$ be the reflection of $Q$ over $M$. Analogously, we get that $Q'$ is the $A$-excenter of triangle $AMC$.
Let $R'$ be the intersection of lines $BP'$ and $CQ'$. Note that $R'$ is the $A$-excenter of triangle $ABC$. Point $R$ is the reflection of $R'$ over $M$ as by reflecting lines $BP'$, $CQ'$ over $M$ we get lines $CP$, $BQ$. It is well-known that distance of incenter and the distance of $A$-excenter from the perpendicular bisector of $BC$ is the same, thus by reflecting $R'$ over $M$ we get a point ($R$) that lies on the line through $I$ perpendicular to $BC$.

Comment. Another way to finish the problem is by using $BP' \parallel CP$ and $CQ' \parallel BQ$ to show that $I$ is the orthocenter of triangle $RBC$, as $IB \perp BP'$ and $IC \perp CQ'$.

Solution:
Let $A'$ and $J'$ be the reflections of $A$ and $J$ in $M$, and let us denote $\angle ABC$ by $\beta$.
First we show that $A', J'$ and $P$ are collinear. Obviously, $\angle JMP = 90^\circ$, which implies that $\angle PJJ' = \angle PJ'J$. Simple angle chasing shows that $\angle MJA = 90^\circ + \beta / 2$. Since $\angle AJP = \beta$, we obtain $\angle MJP = 90^\circ - \beta / 2$, and by the reflection it follows that $\angle MJ'A' = 90^\circ + \beta / 2$ and $\angle MJ'P = 90^\circ - \beta / 2$. This yields that $A', J'$, and $P$ are collinear.
Now we observe that $P$ is the intersection of two angle bisectors, thus $P$ is the $A'$-excenter of $MA'C$. It follows that $BI$ and $CP$ are two perpendicular angle bisectors. One can show similarly that $CI \perp BQ$, hence $I$ is the orthocenter of $BCR$, and the statement follows.

Solution:
It is enough to show that $I$ is the orthocenter of triangle $RBC$. By symmetry, it suffices to prove that $BI \perp RC$. Denote the midpoints of sides $AB, AC$ by $B'$ and $C'$, respectively. The famous Iran lemma tells us that the projection of $C$ onto line $BI$ lies on $MC'$. Thus, we need to prove that lines $BI, MC', CP$ are concurrent. This is the same as saying that triangles $BCC', JPM$ are perspective, which - by Desargues's theorem - is equivalent to the points $BC \cap JP, CC' \cap PM, C'B \cap MJ$ being collinear.
Now let us define some new points. Let the circumcircle of $AJB$ intersect lines $AM, BM$ for the second time at $S$ and $T$, respectively. Since this circle is symmetric with respect to $MJ$, and so are the lines $AM, BM$, we have that $ASBT$ is an isosceles trapezoid. Notice that the angle condition of the problem tells us
$$\angle AJT + \angle AJP = \angle ABT + \angle AJP = \angle ABT + \angle ABC = 180^\circ,$$
thus $T, J, P$ are collinear. Now observe that $\angle JMK = \angle JMA + \angle KMA = \angle BMA / 2 + \angle CMA / 2 = 180^\circ / 2 = 90^\circ$. Also, as $MJ$ is the perpendicular bisector of $AT$ and $BS$, we have that lines $MK, AT, BS$ are all perpendicular to $MJ$, so they are all parallel.
Let $U$ be the midpoint of $AT$ and $V$ be the point at infinity of line $MK$. As we saw previously, $U$ lies on $MJ$ and $V$ lies on $AT$. Now, $BC \cap JP = BC \cap UV$, $CC' \cap PM = CC' \cap VM$, $C'B \cap MJ = C'B \cap MU$. We wish to show that these points are collinear, which now becomes same as saying that triangles $BCC', UVM$ are perspective. By Desargues's theorem, it suffices to prove that lines $BU, CV, C'M$ are concurrent.
Suppose that lines $BU, C'M$ intersect at $X$. Notice that $U, B', C'$ all lie on a midline of triangle $ABC$ and $B'B \parallel C'M$. It follows that triangles $BMX, UB'B$ are homothetic, so it is enough to prove our goal for triangle $UB'B$. That is, we need to show that if $U'$ is the reflection of $U$ in $B'$, then $BU'$ is parallel to $BS$. However, this is fairly trivial, as $B'$ lies on the midline of the isosceles trapezoid $ASBT$ (since it is the midpoint of diagonal $AB$), so $U'$ must lie on $BS$. We are finally done.