Let $KH$ intersect $SR$ at $X$ .It is easy to observe that $QX \perp SR$ (simpson's line ).But $\angle PSR =90 \Longrightarrow KSXQ$ is a rectangle .Hence $KH$ bisects $QS$ as desired

Furthermore, if we have a $\triangle{ABC}$, and $D$ a point on its circumcircle, then $D$-Simson line passes through the midpoint of$DH$, where $H$ is the ortochenter of $\triangle{ABC}$.

Let HK meet QS at L. Note that points P,Q,H,K are concyclic (Why?)Thus <QKL=x implies <QPH=<QPR=<QSR=x Now <RSP=<QKP=90 degrees imply that RS is parallel to QK. Thus <QSR=<SQK=x. So KL=LQ. Similarly KL=LS and result follows.

By Simson line, $L=\overline{SR}\cap\overline{KH}$ is the foot from $Q$ to $\overline{SR}.$ Since $\angle KSL=90,$ we see $KSLQ$ is a rectangle. $\square$