Let $CX$ and $BY$ meet $(ABCDEF)$ again at $G$ and $H$ respectively. Let $AB\cap DC=S$ and $AG\cap DH=T$. By Pascal on $BHDCGA$, $T,P,S$ are collinear. By Pascal on $BAECDF$, $Q,S,O$ are collinear. Note that $YB=YF$ implies $DH\parallel BF$, so $DT\parallel FQ$. Similarly, $AT\parallel EQ$. We also have $AD\parallel EF$, so there is a homothety sending triangle $TAD$ to $QEF$. This means that the lines $TQ,AE,DF$ are concurrent, so $T,O,Q$ are collinear. Combining everything, $QSPOT$ is collinear, so we are done.

Complex bash with $ABCDEF$ inscribed in the unit circle, so that $$|a|=|b|=|c|=|d|=|e|=|f|=1$$$$ad = ef$$$$\overline{x} = \frac{x}{ce} = \frac{a+e-x}{ae} \implies x = \frac{c(a+e)}{a+c}$$$$\overline{y} = \frac{y}{bf} = \frac{d+f-y}{df} \implies y = \frac{b(d+f)}{b+d}$$$$o = \frac{ae(d+f) - df(a+e)}{ae-df}$$$$q = \frac{bf(c+e) - ce(b+f)}{bf-ce}$$Let lines $CX,BY$ intersect the unit circle again at $R,S$ so that $$r = \frac{c-x}{c\overline{x}-1} = \frac{c-\frac{c(a+e)}{a+c}}{\frac{c(a+e)}{e(a+c)}-1} = \frac{ce(c-e)}{a(c-e)} = \frac{ce}a$$$$s = \frac{b-y}{b\overline{y}-1} = \frac{b-\frac{b(d+f)}{b+d}}{\frac{b(d+f)}{f(b+d)}-1} = \frac{bf(b-f)}{d(b-f)} = \frac{bf}d$$$$p = \frac{cr(b+s) - bs(c+r)}{cr-bs} = \frac{\frac{bc^2e}a+\frac{bc^2ef}{ad}-\frac{b^2cf}d-\frac{b^2cef}{ad}}{\frac{c^2e}a-\frac{b^2f}d} = \frac{bc(abf+bef-cde-cef)}{ab^2f-c^2de}$$Now in each of these points we substitute $d = \frac{ef}a$ so that $$o = \frac{ae\left(\frac{ef}a+f\right) - \frac{ef^2}a(a+e)}{ae-\frac{ef^2}a} = \frac{ae^2f+a^2ef-aef^2-e^2f^2}{a^2e-ef^2} = \frac{ef(a+e)(a-f)}{e(a+f)(a-f)} = \frac{f(a+e)}{a+f}$$$$p = \frac{bc\left(abf+bef-\frac{ce^2f}a-cef\right)}{ab^2f-\frac{c^2e^2f}a} = \frac{bc(a^2bf+abef-acef-ce^2f)}{f(a^2b^2-c^2e^2)} = \frac{bcf(a+e)(ab-ce)}{f(ab+ce)(ab-ce)} = \frac{bc(a+e)}{ab+ce}$$Then we compute $$p-o = \frac{(a+e)\left[bc(a+f) - f(ab+ce)\right]}{(a+f)(ab+ce)} = \frac{(a+e)(abc-abf+bcf-cef)}{(a+f)(ab+ce)}$$and \begin{align*} q-o &= \frac{(a+f)(bcf+bef-bce-cef) - f(a+e)(bf-ce)}{(a+f)(bf-ce)} \\ &= \frac{abcf+abef-abce+bcf^2-bcef+ce^2f-cef^2-abf^2}{(a+f)(bf-ce)} \\ &= \frac{(f-e)(abc-abf+bcf-cef)}{(a+f)(bf-ce)} \end{align*}If $abc-abf+bcf-cef = 0$, then $O,P,Q$ are all the same point. Otherwise, we cancel it out and write $$\frac{p-o}{q-o} = \frac{(a+e)(bf-ce)}{(f-e)(ab+ce)}$$which is real. $\blacksquare$