Thanks everyone, for their hints! I think I got the solution (using @cr001 's hint ) . I hope someone can verify this proof .
I am going to use @cr001's diagram.

Let $AC\cap EF= I$ . Let $H_1$ be the foot of the perpendicular from $P$ to $BC$.
Let $H_2$ be the foot of the perpendicular from $Q$ to $DC$. Let $H_3$ be the foot of the perpendicular from $P$ to $BA$.Let $H_4$ be the foot of the perpendicular from $Q$ to $AD$.Let $H_5$ be the foot of the perpendicular from $I$ to $BA$.Let $H_6$ be the foot of the perpendicular from $I$ to $AD$.
Now note that AI is the angle bisector of EF. So we have $\frac {AE}{AF}=\frac {EI}{IF}$ (using the angle bisector theorem )
Also we have $\frac {EP}{EI}=\frac {PH_3}{IH_5}$ (using similarity ).
similarly, we have $\frac {FQ}{FI}=\frac {QH_4}{IH_6}$ (using similarity ).
So we have $\frac {IH_5}{PH_3} \cdot\frac {QH_4}{IH_6}= \frac {EI}{EP}\cdot\frac {FQ}{FI}=\frac {AE}{EP}\cdot \frac {FQ}{AF}=1 \implies \frac {IH_5}{PH_3} \cdot\frac {QH_4}{IH_6}=1 \implies QH_4= PH_3$ (since $IH_5=IH_6$).
So we have $DH_2=QH_4= PH_3=BH_1 \implies CH_1=CH_2$ .
Now, since $\angle PH_1B=\angle BXP=90 $, we get $PH_1BX$ cyclic .
Similarly $QYH_2D$ is cyclic.
So $\Bbb P(C,(PH_1BX))= CH_1\cdot CB=CH_2\cdot CD=\Bbb P(C,(PH_1BX))$
So $\Bbb P(C,(PH_1BX))=\Bbb P(C,(QH_2YD)) \implies CX \cdot CP=CY \cdot CQ \implies XYPQ$ is cyclic .
And we are done!