Remark: The problem as stated is not entirely correct. Only when both $\angle ABC$ and $\angle ACB$ are acute angles do we have that $D$ is the incenter of the triangle $IJK$. When $\angle ABC$ or $\angle ACB$ is a right angle, we obtain a degenerate case, where $M=E=D=B$ so that $IJ=AB$, or $N=F=D=C$ so that $IJ=AC$, respectively. If $\angle ABC>\dfrac{\pi}{2}$ or $\angle ACB>\dfrac{\pi}{2}$, then $D$ is an excenter of the triangle $IJK$. See an illustration below. (If we define $I$ and $J$ so that $MI$ and $NJ$ are disjoint, then in the case $\angle ABC>\dfrac{\pi}{2}$, $D$ is the excenter of the triangle $IJK$ opposite to $I$. On the other hand, if $\angle ACB>\dfrac{\pi}{2}$, then $D$ is the excenter of the triangle $IJK$ opposite to $J$.)
Let $R$ denote the circumradius of the triangle $ABC$. Write $\alpha$, $\beta$, and $\gamma$ for the angles $\angle BAC$, $\angle CBA$, and $\angle ACB$, respectively. In what follows, we assume that $\beta$ and $\gamma$ are in the interval $\left(0,\dfrac{\pi}{2}\right)$. First, we assume that $\alpha\in\left(0,\dfrac{\pi}{2}\right)$. Then, $$\angle DFE=\pi-2\gamma\,.$$ Thus, the distance from $D$ to $EF$ is given by $$DF\,\sin(\pi-2\gamma)=DF\,\sin(2\gamma)\,.$$
Since the triangle $ABC$ and the triangle $DBF$ are similar, we get
$$DF=AC\,\left(\frac{BD}{AB}\right)=AC\,\cos(\beta)=2\,R\,\sin(\beta)\,\cos(\beta)=R\,\sin(2\beta)\,.$$
Thus, the distance from $D$ to $EF$ is $R\,\sin(2\beta)\,\sin(2\gamma)$, whence the distance from $D$ to $IJ$ is
$$\rho:=\frac{1}{2}\,R\,\sin(2\beta)\,\sin(2\gamma)\,.$$
It is not difficult to see that $OA\perp EF$, whence $OA \perp IJ$. Therefore, $A$ is the midpoint of the arc $IAJ$. This shows that $$\angle DKI=\angle AKI=\angle AKJ=\angle DKJ=:\theta\,.$$ Ergo, the distance between $O$ and $EF$ is
$$\begin{align}d:=OA-AE\,\sin(\beta)&=R-AB\,\left(\frac{AE}{AB}\right)\,\sin(\beta)\\&=R-\big(2\,R\,\sin(\gamma)\big)\,\cos(\alpha)\,\sin(\beta)\\&=R\big(1-2\,\cos(\alpha)\,\sin(\beta)\,\sin(\gamma)\big)\,.\end{align}$$
Thus, the distance $\delta$ between $O$ and $IJ$ is
$$\begin{align}\delta=d-\rho&=R\,\Big(1-2\,\cos(\alpha)\,\sin(\beta)\,\sin(\gamma)-2\,\cos(\beta)\,\cos(\gamma)\,\sin(\beta)\,\sin(\gamma)\Big)
\\
&=R\,\Big(1+2\,\big(\cos(\beta+\gamma)-\cos(\beta)\,\cos(\gamma)\big)\,\sin(\beta)\,\sin(\gamma)\Big)
\\
&=R\,\Big(1-2\,\sin^2(\beta)\,\sin^2(\gamma)\Big)\,.\end{align}$$
Finally, note that $$\begin{align}DK=DH&=BD\,\cot(\gamma)=\big(AB\,\cos(\beta)\big)\,\cot(\gamma)\\&=\big(2\,R\,\sin(\gamma)\big)\,\cos(\beta)\,\cot(\gamma)\\&=2\,R\,\cos(\beta)\,\cos(\gamma)\,.\end{align}$$
Since $\delta=R\,\cos(2\,\theta)=R\,\big(1-2\,\sin^2(\theta)\big)$, we conclude that
$$\sin(\theta)=\sin(\beta)\,\sin(\gamma)\,,$$
whence the distance from $D$ to $IK$ or to $JK$ is
$$DK\,\sin(\theta)=DK\,\sin(\beta)\,\sin(\gamma)=2\,R\,\cos(\beta)\,\cos(\gamma)\,\sin(\beta)\,\sin(\gamma)=\rho\,.$$ Hence, $D$ is the incenter of the triangle $IJK$, with inradius $\rho=\dfrac{1}{2}\,R\,\sin(2\beta)\,\sin(2\gamma)$.
The case $\alpha \in\left(\dfrac{\pi}{2},\pi\right)$ can be dealt with in a similar manner. One of the differences is that $\angle DFE=2\gamma$.
Best Answer
I will give a proof for the problem in the OP by also adding some bonus properties. The given geometric constellation can be embedded in the following one, isolated in a new statement below that uses the same notation from the OP for a simple parallel (against my will). The bonus properties come according to my principle that if two lines are perpendicular, and if the point "has a meaning", then a faithful proof should elucidate, or at least mention this. Full details are given, this makes the exposition longer. For a short exposition stop after the statement. (Which is longer than the proof, and fill in the details by a glance.)
Proof:
(1) In the Euler circle we have $EF\perp MY$, since $MF=MB=MC=ME$ and $YF=YH=YA=YE$ (in a right triangle the median from the right angle is half hypotenuse), so $\Delta EMY=\Delta FMY$ so the chord $EF$ is perpendicular on the diameter $MY$ in $(9)$.
Then we have $EF\|AP$, since $EF$ is an "anti-parallel" to $BC$, using (blue) angles explicitly, $\widehat{PAB}=\frac 12\overset\frown{AB}=\hat C=\widehat{EFA}$.
And of course $PA\perp AO$ by construction.
(2) This is again simple. By (1) we have $MY\perp AP$, and of course $PBDMC$. So $Y$ is orthocenter in $\Delta APM$. The third height $AY$ intersects the side $AM$ in a point $X'$ such thet $\widehat YXM= 90^\circ$, so $X'=X$, since $YM$ is diameter in $(9)$.
(3) To be able to focus only on the few points involved, here is an other picture:
To have a simple proof, we reverse slightly the order of the construction of the points, with reference to the above picture. We construct first the points $A,B,C;D,E,F;H$ as above. Then we draw the circle $(AEFH)$ with diameter $AH$, and denote by $\Xi,\Upsilon$ the intersections of this circle with the lines $ZA$, $ZH$, respectively. Let $A\Upsilon$ intersect the line $PZBC$ in a point $M'$. (We want $M'=M$, so let us show it.) The using the power of the point $Z$ w.r.t. different circles, we have: $$ \begin{aligned} ZD\cdot ZM' &=ZH\cdot Z\Upsilon &&\text{ powers of $Z$ w.r.t. the circle }(M'DHY)\\ &=ZE\cdot ZF &&\text{ powers of $Z$ w.r.t. the circle }(AFHE)\\ &=ZD\cdot ZM &&\text{ powers of $Z$ w.r.t. the circle }(9)\ . \end{aligned} $$ In a computer game, $M$ would inherit the property of $M'$, so $AH\Upsilon\perp AM$. Given that $AHD\perp AM$, the point $H$ is the orthocenter in $\Delta AZM$, and thus the third height $MH$ goes through the projection of $H$ on $AZ$, which is $\Xi$.
(4) We first observe that $APZQ$ is a parallelogram, $AQ\| PZ$ by construction, and $EF\|AP$. (Or use angle chasing to show that (two of) the "red angles" in $P,Q,Z$ have the measure $\hat B-\hat C$.)
The two triangles questioned about similarity have each a right angle in $A,D$, so it is enought to show a proportionality of two pairs of sides. We take the easy ones, and want to show (!): $$ \begin{aligned} \frac{AQ}{PD} &\ \overset{(!)}= \ \frac{AY}{YD}\ . &&\text{ Equivalently:} \\ \frac{PZ}{PD} &\ \overset{(!)}= \ \frac{AY}{YD}\ . &&\text{ Equivalently, using derived proportions:} \\ \frac{PZ}{PD-PZ} &\ \overset{(!)}= \ \frac{AY}{YD-AY}\ . &&\text{ Equivalently:} \\ \frac{PZ}{ZD} &\ \overset{(!)}= \ \frac{YH}{HD}\ . &&\text{ True, because of }ZH\|PY\ . \end{aligned} $$ (Since $ZH$, $PY$ are perpendicular on $AM$ in $\Upsilon$, respectively $X$.)
The two angles in $Y$ of the two similar triangles are equal, given that $A,Y,D$ are colinear, we obtain $Q,Y,P$ colinear. And we already know $X$ is on the same line (with $P,Y$). The final touch is $$ AXM\perp AYXQ\text{ in }X\ . $$
$\square$