Let $E$ be a point on $PQ$ such that $QE=PC$. Then $EP=CB=SD$, so $SDPE$ is a parallelogram, so $SE\parallel DP$. Also $RS\parallel AP$, thus $S,E,R$ are collinear. Thus $\angle SRQ=\angle PAC=\angle DBC=\angle SCQ$, so $SCRQ$ is cyclic.

oneplusone wrote: [...] Also $RS\parallel AP$[...] Why?

Because the midpoints of $AR,CQ,PQ$ are the same, so $APRE$ is a parallelogram.

We need to prove that $SR//AD$,so let $M$ be the reflection of $A$ wrt $P$,so we have $MB=AQ=CR$,so $CBMR$ is a parallelogram so $BC=MR=SD$,so $SDMR$ is a parallelogram,so we are done.