Let $H$ be the orthocenter of let $AH$ intersect $BC$ at $A'$. Note that $BP^2=BC'\cdot BA=BQ^2$ so $BQ=BP$. Let $\Omega$ be the circle with center $B$ and radius $r=BP=BQ$. Inversion in $\Omega$ swaps $(C',A)$, $(H, B')$ and $(A',C)$. Let $K$ be the intersection of segment $AA'$ with $\Omega$. Since $A$, $K$, $H$ and $A'$ are colinear, $BCB'KC'$ is cyclic. Also since $\angle BA'K=90^{\circ}$, circles $(BA'K)$ and $\Omega$ are tangent to each other. This implies that $CK$ is tangent to $\Omega$. Now we have $\angle KC'H=\angle KC'C=\angle KBC=\angle KBP=\angle A'KC$ so $CK$ is also tangent to $(C'HK)$. This gives $$CA\cdot CB'=CC'\cdot CH=CK^2=CP\cdot CQ$$or equivalently, $AB'PQ$ is cyclic.

Note that $BP=BQ=\sqrt{BC'\cdot BA}$, so \begin{align*} CP\cdot CQ&=(BC+BP)\cdot(BC-BP)\\ &=BC^2-BP^2\\ &=BC^2-BC'\cdot BA\\ &=BC(BC-BA\cos B)\\ \end{align*}and \begin{align*} CB'\cdot CA&= CA\cdot BC\cos C \end{align*}However, we are done after we note that if $D$ is the foot from $A$ to $BC$, $BA\cos B=D$ and $CA\cos C=CD$, so the two values sum are equal, as desired.