Let $NL$ meet the circle $(NKR)$ at $A.$ $\bold{\underline{Claim:}}$ $SR$ bisects $\angle ARH.$ $\bold{\underline{Proof:}}$ We have $$\angle KRA =\angle KNA =\angle PNM =\angle PRL =\angle R.$$But $\angle QRK =\angle KRP =\frac{\angle R}{2}.$ Therefore, $\angle ARS = \frac{\angle R}{2}.$ This prove our claim. $\circ$ Now, $SR \perp HA$ and $SR$ bisects $\angle ARH.$ Therefore, $SR$ is the perpendicular bisector of $HA.$ So, $\boxed{SH=SA}.$ Also, $\angle SKA =\angle SRA =\angle SRK =\angle SAK.$ Therefore, $\boxed{SK=SA}.$ Hence, $\boxed{SH=SK},$ i.e. $\Delta SHK$ is an isosceles triangle. $QED.$

[asy][asy] import olympiad; import geometry; pair P, Q, R, I, K, N, M, S, A, L, H, X; R=dir(-20); P=dir(120); Q=dir(200); X=orthocenter(K,P,R); I=incenter(P, Q, R); L=foot(P, Q, R); M=foot(X, R, P); H=intersectionpoint(line(P, L), line(K, R)); N=intersectionpoint(line(K, M), line(P, L)); K=intersectionpoint(line(R, I), line(P, Q)); draw(P--L); draw(K--M); draw(K--R); draw(circumcircle(K, N, R), orange); draw(P--Q--R--cycle); label("R", dir(-20), NE); label("P", dir(120), NW); label("Q", dir(200), W); [/asy][/asy]

Let $\angle QLK=\angle KRP=\theta$. Clearly, $\angle LHR=\angle NHK=90^{\circ}-\theta$, and $\angle RKM=90^{\circ}-\theta$. Thus, we have that $\triangle NHK$ is $N$-isosceles. Now, $\angle SRK=\angle SNK=\theta=\angle SNH$, giving us that $NS$ is the perpendicular bisector of $KH$. Thus we have that $SK=SH$, and we are done!