Not that hard to prove

Let $X$ be the midpoint of big arc $BC$, and $Y=AX\cap IP$. $\angle EIY=\angle BMX=\angle XAB$, so we get $(I,E,A,Y)$ is cyclic, and $\Delta AIY\sim BXM$. And $\angle PAY=\angle PMX=\angle XPN$. So $IP:PY=MN:NX$, and we can get $\Delta EIP\sim \Delta BMN$

Let $X$ be the center of spiral similarity sending $EI$ to $FJ.$ We will assume that $E$ is on segment $AB$ and $F$ is on the extension of $AB$ past $B.$ Let $E'$ be the point on segment $AC$ for which $CE' = CD.$ Then it's easily seen that $AEIE'$ is cyclic since $\angle EIE' = 360 - 2 \angle BIE - 2 \angle CIE' = 360 - 2 (\angle BIC) = 180 - \angle A.$ Let $\Gamma = (\triangle ABC), \gamma = (AEIE').$ Let $M_a = AI \cap \Gamma.$ Let $A'$ be the point opposite $A$ on $\Gamma$ and let $P_1, P_2$ denote $IA' \cap \Gamma, JA' \cap \Gamma.$ We claim that $P_1, P_2$ are the two desired points $P$ and $Q,$ respectively. By spiral similarity, it only suffices to show that the spiral similarity sending $EI$ to $FJ$ about $X$ also sends $P_1$ to $P_2.$ Notice that since $X$ sends $EI$ to $BM_a$ (the midpoints), we also have that $X \in (\triangle ABM_a) = \Gamma.$ Furthermore, since $X = (AEE') \cap (ABC),$ we know that $\frac{BX}{XC} = \frac{BE}{CE'} = \frac{BD}{DC},$ which hence implies that $X, D, M_a$ are collinear by the Angle-Bisector theorem in $\triangle BXC.$ To show that $P_1, P_2$ work, we only need to show that $\triangle XP_1I \sim \triangle XP_2J$ and $P_1P_2 || BC.$ Note that $\angle XP_1I = \angle XP_1A' = \angle XP_2A' = \angle XP_2J.$ Now, in order to show that $\angle XIP_1 = \angle XJP_2,$ we just need $180 - \angle XIA' = \angle XJA',$ i.e., $XIA'J$ is cyclic. To show this, let's invert with respect to $M_a$ with radius $M_aB.$ Then, $X, I, A', J$ are mapped to $D, I, M_aA' \cap BC, J,$ respectively. Let $Y = M_aA' \cap BC, Z = AI \cap BC.$ Then, observe that $\angle ADY = 90 = \angle AM_aA' = \angle AM_aY,$ and so hence $ADM_aY$ is cyclic. Therefore, by PoP at $Z,$ we know that $DZ * ZY = AZ * ZM_a = BZ * ZC = IZ * ZJ,$ where the last part follows since $BICJ$ is cyclic. But $DZ * ZY = IZ * ZJ \Rightarrow DIYJ$ is cyclic, and so inverting back at $M_a$ gives the conclusion that $\angle XIP_1 = \angle XJP_2.$ By the two previous paragraphs, we know now that $\triangle XP_1I \sim \triangle XP_2J.$ Finally, let's show that $P_1P_2 || BC.$ To do so, it suffices to prove that $A'I, A'J$ are isogonal conjugates w.r.t. $\angle BA'C.$ But this follows immediately by sin Ceva in $\triangle BA'C$ with the points $I, J.$ $\square$