Let $T$ be the midpoint of $BC$, $S$ be the antipode of $A$ wrt. to $(ABC)$ and let $CH$ intersects $(ABC)$ again at $X$. It's well-known that $H,T,S$ collinear. Homothety center at $C$ with ratio $2$ gives us $N',H,T,S$ collinear where $HT\cap (ABC)=N'\neq S$. Also, $\angle{TN'C}=90^{\circ}$. Hence $N'\equiv N$. Note that $TM^2-MH^2=TF^2+(FM-MH)(FM+MH)=TF^2+FX\times FC=TF^2+FA\times FB=FT^2+(TB^2-FT^2)=TB^2$. Hence the power of $T$ respect to $(M,MH)$ is equal to $TB^2$. This gives $TH\times TN=TB^2$. So $\angle{TNP}=\angle{HBT}=\angle{HCA}$, which means $C,N,H,P$ concyclic. This gives $\angle{HPC}=90^{\circ}$. Let $HP\cap BC=B'$. We get that the reflection of $B'P$ over the $C$-altitude of $\triangle{ABC}$ is the $B$-altitude. Reflect $Q$ over $F$ to get point $Q'$. Note that $X$ is the reflection of $H$ over $F$ and we also know that $B$ is the reflection of $B'$ over $F$. Hence $X,Q',B$ collinear. Also, note that $F$ is the intersection of chords $AB$ and $CX$. And there's another chord $l$ (which is $QQ'$) that pass through $F$ and intersect $AB,CX$ at $Q,Q'$. Moreover, $FQ=FQ'$. By converse of Butterfly's, we get that $OF\perp l$, which means $OF\perp QQ'$. Hence $\angle{OFQ}=\frac{180^{\circ}}{2}=90^{\circ}\Rightarrow QF\perp OF$, done.

Let $\Gamma$ denote the circumcircle of $\triangle ABC$. Let $\overline{NH}$ and $\overline{CD}$ meet $\Gamma$ again at $X$ and $Y$, respectively. It is well-known that $\overline{CX}$ is a diameter of $\Gamma$. Now we observe $$\measuredangle BNX=\measuredangle FCA=\measuredangle HCP$$so $C,N,H,P$ are concyclic. By Reim's Theorem, $\overline{HP} \parallel \overline{XY}$. Claim. Let $E$ be the reflection of $B$ in $\overline{CF}$. Then $E$ lies on line $\overline{HQ}$. (Proof) Angle Chase! (since $\overline{HQ} \parallel \overline{XY}$ all required angles are easy to find). $\blacksquare$ Reflect $Q$ about $F$ to get $R$ and extend $\overline{FC}$ to meet $\Gamma$ again at $H_C$. By converse of Butterfly's Theorem for chords $\overline{CH_C}, \overline{AB}, \overline{QR}$, we see $\overline{OF} \perp \overline{QR}$, concluding the proof. $\blacksquare$ Comment. Excluding the bit about Reim's Theorem, this is essentially Balkan 2008/1.

Let $J$ be the midpoint of $AB$. We know that $MH\|=OJ$ so $HJ\| OM$ so $H,J,N$ are collinear and also $N\in (O)$. Let $P'$ be the perpendicular foot of $H$ on $AD$, so $C,N,H,P'$ lies on a circle and $\angle CNP'=\angle CHP'=\angle CDA=\angle CAB=\angle CNB$ so $N,P',B$ are collinear and $P\equiv P'$. Let $AD,BE$ be the attitude of triangle $ABC$. Let the line through $H$ and parell to $BC$ intersect $DE$ at $X$. In this post (https://artofproblemsolving.com/community/u479101h1848465p12465948), I have pointed out that $\angle XFO=90^\circ$. We have to prove that $X,Q,F$ are collinear. Let $CF\cap ED=S$. It's well known that: $(AH,SF)=-1$. Let $XH\cap AC=T$. We have $H$ be the orthocentre of $AQI$ and $\angle THQ=\angle HQI=\angle HAT=\angle EHT$ so $HT$ is the bisector of angle $EHQ$ so $(CT,EQ)=-1$ that means $X(CH,SQ)=-1$. Otherwise $X(CH,SF)=-1$. And this leads to $X,Q,F$ are collinear so we have Q.E.D

Here is a complex numbers bashing solution. Denote by $x$ the affixe of $X$.As usual set $(ABC)$ to be the unit circle so that we have: $$|a|=|b|=|c|=1,o=0,h=a+b+c,f=\frac{1}{2}(a+b+c-ab\overline{c}),e=\frac{1}{2}(a+b+c-a\overline{b} c)$$First,some synthetic observations.Let $E$ the foot of the $B$-altitude. Claim:$P$ is the reflection of $E$ into $HC$ Proof:

We are ready for the bash Let's redefine $Q$ as the point on $AC$ for which $QF\perp OF$,so that we have: $q=a+c-\overline{q}ac$ $\frac{q-f}{f}=-\overline{\frac{q-f}{f}}\iff q=-\frac{f}{\overline f}\overline{q}+2f$. Equating these equations: $\overline q=\frac{a+c-2f}{ac-\frac{f}{\overline{f}}}=\frac{a+c-a-b-c+ab\overline{c}}{ac-\frac{a+b+c-ab\overline{c}}{\frac{ab+bc+ca-c^2}{abc}}}=\frac{b(a\overline{c}-1)\frac{ab+bc+ca-c^2}{abc}}{\frac{ab+bc+ca-c^2}{b}-a-b-c+ab\overline{c}}=\frac{b(a-c)(ab+bc+ca-c^2)\frac{1}{ac^2}}{ab+bc+ca-c^2-ba-b^2-bc+ab^2\overline{c}}=\frac{b(a-c)(ab+bc+ca-c^2)\frac{1}{ac^2}}{c(a-c)+b^2(\frac{a}{c}-1)}=\frac{b(ab+bc+ca-c^2)}{(b^2+c^2)ac} \iff q=\frac{\frac{1}{b}(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}-\frac{1}{c^2})}{\frac{(b^2+c^2)ac}{b^2c^2}}=\frac{c^2(a+b+c-ab\overline{c})}{b^2+c^2}$. Now,as $P$ is the reflection of $E$ in $HC$ we obtain: $\overline{\frac{e-c}{h-c}}=\frac{p-c}{h-c}\iff p=\frac{h-c}{\overline{h}-\overline{c}}(\overline{e}-\overline{c})+c=\frac{a+b}{2\frac{a+b}{ab}}(\overline{a}+\overline{b}-\overline{c}-\overline{a}b\overline{c})+c=\frac{1}{2}(b+a-ab\overline{c}-b^2\overline{c})+c$. We want to show that $Q,H,P$ are collinear.Note that: $h-q=\frac{-c^2(a+b+c-ab\overline{c})+(b^2+c^2)(a+b+c)}{b^2+c^2}=\frac{-c^2a-c^2b-c^3+abc+b^2a+b^3+b^2c+c^2a+c^2b+c^3}{b^2+c^2}=\frac{abc+b^2a+b^3+b^2c}{b^2+c^2}=\frac{b(b+c)(b+a)}{b^2+c^2}$, and $2(h-p)=-a-b+\frac{ab}{c}+\frac{b^2}{c}+2c-2a-2b-2c=\frac{ab+b^2}{c}+a+b=\frac{(b+a)(b+c)}{c}$. Finally,we obtain:$\frac{h-q}{h-p}=\frac{2bc}{b^2+c^2}=\frac{\frac{2}{bc}}{\frac{b^2+c^2}{b^2c^2}}=\overline{\frac{2bc}{b^2+c^2}} \iff \frac{h-q}{h-p}\in\mathbb{R}\iff Q,H,P$ are collinear $\square$