We use Cartesian coordinates: WLOG, let $A$ be at $(0,0)$, $B$ at $(0,1)$, $C$ at $(1,1)$, and $D$ at $(1,0)$. Now, let $P$ be at $(k,1)$. Rotating $P$ 90 degrees clockwise around $A$ gives that $S$ is at $(1,-k)$. From this, we have that $R$ is at $(1+k,1-k)$, so the slope of $CR$ is $-1$. Thus, $CR$ is perpendicular to $AC$, a diameter of the circumcircle of $ABC$, and is therefore tangent to the circle.

Let $O$ be the center of the square $ABCD$ It is enough to prove that $\angle DYR = \angle ODR/2 = 90^o /2 = 45^o$ (thanks to tangent -chord angle) Let $M$ be the projection of $R$ on line $BY$. Let $BD=y$ and $DC = x$, so $AB= BC =x+y$ Right triangles $ABD$ are congruent so $RM=BP=y$ and $PM= AB = x+y$ so $MC = PM - PC = (x+y) - x= y = RM$ , which means that Right triangle $RCM$ is also isosceles so $\angle RCM= 45^o = \angle DCR$