Let $AA'$ be the altitude, noticed that $$(P, Q ; A', C)=-1$$So $$CP \times CQ =CA' \times CB =CA' \times CB$$As desired

Notice that $$CP \cdot CQ=BC^2-BP^2=BC^2-BA\cdot BC'=BC^2-BC\cdot BA'=CB\cdot CA'=CB'\cdot CA,$$hence $A, B', P, Q$ are concyclic.

An alternative solution is also possible: Invert at $A$ with power $\sqrt{AH\cdot AA'}$, where $H$ is the orthocenter of $ABC$. The desired result is equivalent to $C$ lying on the polar of $B$ in $\odot(AH)$; which is just an application of Brokard's Theorem for the cyclic quadrilateral $AB'HC'$.

Triple posting but another solution WizardMath wrote: 10.5 Let $BB'$, $CC'$ be the altitudes of an acute triangle $ABC$. Two circles through $A$ and $C'$ are tangent to $BC$ at points $P$ and $Q$. Prove that $A, B', P, Q$ are concyclic. Let $H$ be the orthocenter of $\triangle ABC$. Reflect $C'$ in $B$ to get point $C_1$ and $H$ in $A'$ to get point $H_1$. Note that $C'$ is the HM Point opposite $A$ in $\triangle APQ$. As $\measuredangle AC'H=90^{\circ}$, it follows that $H$ is the orthocenter of $\triangle APQ$ as well. Notice that $B'$ is the spiral center for $\mathcal{S}: \overline{HA'} \mapsto \overline{C'B}$. As $\mathcal{S}: H_1 \mapsto C_1$, we conclude that $B' \in (AC_1H_1)$. Consequently, $B' \in (APQ)$ and we are done. $\blacksquare$

This follows from my solution here: IMOSL 2008 G4