I assume you mean the projections of $E,F$. If so, since $EFQR$ is cyclic of diameter $EF$, it suffices to probe that the two angles at the base of the quadrilateral are equal. Indeed $\measuredangle QFE=\measuredangle CFB=\measuredangle CPB=\measuredangle DPA=\measuredangle DEA=\measuredangle REF$, as wanted.

Angle chase gives $PE= PF$ and $\angle PER = \angle QFP$, implying that $\triangle PER \stackrel{\sim}{=} \triangle PFQ$, hence $EFQR$ is an isosceles trapezoid.

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mshtand1 wrote: Ahahahahaha, MS_Kekas, nice jokes about legend and masterpiece they are true ; ; $~$ ;

Cute problem. By angle chase, $\measuredangle PCD = \measuredangle FCP$ and $\measuredangle PDC = \measuredangle EDP$. If $W$ is the projection of $P$ onto $CD$, it is easy to see that $PR=PW=PQ$. Another angle chase shows that $\measuredangle DEF = \measuredangle EFC$. Now, if $CF || DE$, the desired statement is obvious. Otherwise, denote by $X$ the intersection point of $CF$ and $DE$ and observe that both lines are perpendicular to the bisector of $\angle CXD$.