Let $QH$ and meet $w$ again at $R$ and let $PR$ meet again at $M'$. Let $MM'$ meet $QR$ at $H'$. Then $H'$ lies on the polar of $P$ with $w$ or the line $AH$ and therefore $H=H'$. This means $M'$ must be the $M$ antipode. Then quadrilateral $AMBM'$ is harmonic so by involution at $P$ quadrilateral $AQCR$ is harmonic, finishing the problem.

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Complex bash with $\omega$ as the unit circle. (I'm assuming $O$ is supposed to be the center of $\omega$.) Then $$|b|=|m|=1$$$$a = \frac{m^2}b$$$$c = -b$$$$o = 0$$$$p = \frac{a^2(b+c) - bc(a+a)}{a^2-bc} = \frac{2b^3m^2}{b^4+m^4}$$$$q = \frac{m-p}{m\overline{p}-1} = \frac{m-\frac{2b^3m^2}{b^4+m^4}}{\frac{2bm^3}{b^4+m^4}-1} = \frac{m(-b^3+b^2m+bm^2+m^3)}{b^3+b^2m+bm^2-m^3}$$$$x = \frac{2ac}{a+c} = \frac{2bm^2}{b^2-m^2}$$Now the circumcenter of $\triangle ABO$ has coordinate $\frac{ab}{a+b}$, so its orthocenter $H$ has coordinate $$a + b + o - \frac{2ab}{a+b} = \frac{a^2+b^2}{a+b} = \frac{b^4+m^4}{b(b^2+m^2)}$$Now we have the vectors $$x-h = \frac{2b^2m^2(b^2+m^2) - (b^2-m^2)(b^4+m^4)}{b(b^2-m^2)(b^2+m^2)} = \frac{-b^6+3b^4m^2+b^2m^4+m^6}{b(b^2-m^2)(b^2+m^2)}$$$$x-q = \frac{2bm^2}{b^2-m^2} - \frac{m(-b^3+b^2m+bm^2+m^3)}{b^3+b^2m+bm^2-m^3} = \frac{m(b^5+b^4m+2b^2m^3-bm^4+m^5)}{(b^2-m^2)(b^3+b^2m+bm^2-m^3)}$$We factor $$x-h = \frac{(-b^3+b^2m+bm^2+m^3)(b^3+b^2m-bm^2+m^3)}{b(b^2-m^2)(b^2+m^2)}$$and $$x-q = \frac{m(b^2+m^2)(b^3+b^2m-bm^2+m^3)}{(b^2-m^2)(b^3+b^2m+bm^2-m^3)}$$So dividing and cancelling gives $$\frac{x-h}{x-q} = \frac{(-b^3+b^2m+bm^2+m^3)(b^3+b^2m+bm^2-m^3)}{bm(b^2+m^2)^2} = \frac{|b^3+b^2m+bm^2-m^3|^2}{|b^2+m^2|^2}$$which is real. $\blacksquare$

sami1618 wrote: so by involution I believe you meant inversion

inversion is an involution on the circle (which preserves cross ratio) Theorem 1.5 here

Oh, okay, you are right