call $ E: AB \cap CD$ so $ \angle BPC = \angle BCA' - \angle CBD = \angle ADB - \angle CBD =$ $ = 90 - \angle ABD - \angle CBD = 90 - \angle ABC = \angle BEC$. Then BCPE is cyclic and $ EB \perp BD$. Then the problem is equal to this: H is the orthocenter of ABC and AH,BH,CH meet respectively BC,CA,AB on D,E,F. Call H' the symmetric of H w.r.t. ED and ED meet AF on G and AH' on K. Then $ KF \perp ED$. We can prove it in this way: F is the harmonic cojugate of G w.r.t. AH, but G on $ \triangle AKH$ is the feet of the internal angle bisector of $ \angle AKH$, so F is the feet of the external angle bisector of $ \angle AKH$ and then $ FK \perp KG$.

Denote AC meet BD at G, B.G.D.P is harmonic division GC is bisector of angle BQC

not a hard one Hint : $AC \cap BD=T$ and $BB' \cap AC=L$ . now use Menelaus theorem for$\bigtriangleup BB'D$ and point $Q$ and use sin law for $\bigtriangleup BCD$ and point $P$ then you will see that $\frac {PB}{PD}=\frac {QB'}{QD}=\frac {BT}{TD}$ so done.

By symmetry, $\Delta AB'Q=\Delta ABQ\implies \angle BQT=\angle DQT$, so we need to show that $\frac{BP}{PD}=\frac{BT}{TD}$, where $T=AC\cap BD$. We see easily that $D$ is the incenter of $\Delta ACP$, so $D, B$ are harmonic conjugate w.r.t. $\overline{PT}$, hence $\frac{BT}{BP}=frac{TD}{PD}\iff\frac{PD}{PB}=\frac{TD}{BT}$; since $TQ$ is internal angle bisector of $\angle BQD$, it follows that $PQ$ is the external bisector of $\angle BQD$ and $PQ\bot AC$. Best regards, sunken rock

Let $AC\cap BD= E $ and $BB'\cap AC = M $. Also, let $BB'\cap PQ = X $. It is easy to see that $A'$ lies on $\odot (ABCD) $. As, $BA = BA'$, so, $BC $ bisects $\angle PCQ $. Thus, $(-1) = (P,E;B,D)\stackrel{Q}{\doublebarwedge}(P,E;B,D)\stackrel{Q}{\doublebarwedge}(X,M;B,B')$. But, $M$ is the midpoint of $BB'$. Hence, $X $ is a point at infinity and so, $BB' || PQ $. This gives $\angle PQC = 90^{\circ} $.

Let $\{L\}$=$BA'\cap CD$, $\{T\}$=$AC\cap BD$, $\{Q\}$=$BC \cap A'D$. Simple angle chasing gives $T,Q,L$ colliniar and $LQ \perp BD$. By a well, known lemma, we get that $QT$ is the polar of $P$, so $P$ is on the polar of $T$ which gives $(P,T,B,D)=-1$. Now, all we have to prove is that $\frac{BT}{TD}=\frac{QB'}{B'D}$, which is Menelaus' theorem in triangle $BB'D$ with line $QT$ $\blacksquare$

Well though the computations took a while They were pretty straightforward and simple And ended nicely I blindly used complex numbers and solved it.