Let $M$ be the midpoint of $AC$. Let $K$ be the projection of $M$ on $AA_1$. As $\angle CA_1A=90^\circ=\angle MKA$ we have $A_1C\parallel MK$. So, $K$ is midpoint of $AA_1$. So, $PK\parallel BA$. So $$\angle PKQ=90^\circ-\angle PKA_1=90^\circ- \angle BAA_1=\angle C_1HA$$. As $\angle CC_1A=\angle CA_1A=90^\circ$ it follows $A_1,C_1,C,A$ lie on a circle. So, we have $\triangle CHA\sim\triangle A_1HC_1$. So, $$\angle PHC_1=\angle MHA=\angle QHA.$$Hence, $$\angle PHQ=\angle C_1HA=\angle PKQ.$$Thus, $P,H,K,Q$ are concyclic. Hence, $\angle QPH=180^\circ-\angle HKQ=90^\circ$. $\blacksquare$ [asy][asy] import olympiad; size(300); pair A,B,C,M,A1,C1,K,Q,P,H; A=(12,0); B=(4,8); C=origin; draw(A--B--C--cycle); M=(A+C)/2; C1=foot(C,A,B); A1=foot(A,B,C); draw(C--C1); draw(A--A1); draw(A1--C1); K=foot(M,A,A1); P=(A1+C1)/2; Q=2K-M; draw(Q--M); draw(Q--P); draw(circumcircle(Q,K,P),red); draw(circumcircle(A,C,A1),blue+dotted); H=orthocenter(A,B,C); draw(P--H); draw(P--K); draw(H--Q); dot("$A$",A,E); dot("$B$",B,N); dot("$C$",C,W); dot("$M$",M,S); dot("$A_1$",A1,NW); dot("$C_1$",C1,NE); dot("$P$",P,N); dot("$K$",K,SE); dot("$Q$",Q,NE); dot("$H$",H,S); [/asy][/asy]