Let $X$ be the second intersection of $AP$ and $(O)$. As $AXBQ$ is an parallelogram, $BQ=AX$. Thus, $AP.BQ=AP.AX$. In the other hand, $AP.AX$ is the power of $A$ wrt $(O)$ which is not depend on $P$.

GorgonMathDota wrote: Given a circle with center $O$, such that $A$ is not on the circumcircle. Let $B$ be the reflection of $A$ with respect to $O$. Now let $P$ be a point on the circumcircle. The line perpendicular to $AP$ through $P$ intersects the circle at $Q$. Prove that $AP \times BQ$ remains constant as $P$ varies. $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ Working in complex numbers, where the circle with center $O$ and radius $OP$ is the unit circle, let $a=a,b=b,p=1,q=q$ and $o=0$ $B$ is the reflection of $A$ with respect to $O:$ $$\Rightarrow b=-a$$$AP\perp PQ$ $$\frac{a-1}{\overline{a}-1}=-\frac{q-1}{\overline{q}-1}$$$$\Rightarrow \frac{a-1}{\overline{a}-1}=q$$$$\Rightarrow a-1=\overline{a}q-q$$$$\Rightarrow \overline{a}=\frac{a+q-1}{q}...(I)$$ $$AP\times BQ=|a-p||b-q|$$$AP\times BQ$ remains constant when $P$ varies if $|a-p||b-q|$ does not depend on $q$ $$|a-p||b-q|=\sqrt{(a-1)(\overline{a}-1)(-a-q)(-\overline{a}-\overline{q})}$$$$\Rightarrow |a-p||b-q|=\sqrt{(a\overline{a}-\overline{a}-a+1)(a\overline{a}+a\overline{q}+q\overline{a}+1)}$$$$\Rightarrow |a-p||b-q|=\sqrt{(a(\frac{a+q-1}{q})-\frac{a+q-1}{q}-a+1)(a(\frac{a+q-1}{q})+\frac{a}{q}+q(\frac{a+q-1}{q})+1)}$$$$\Rightarrow |a-p||b-q|=\sqrt{(\frac{a^2+aq-a-a-q+1}{q}-a+1)(\frac{a^2+aq-a+a}{q}+a+q-1+1)}$$$$\Rightarrow |a-p||b-q|=\sqrt{(\frac{a^2-2a+1}{q})(\frac{a^2+2aq+q^2}{q})}$$$$\Rightarrow |a-p||b-q|=\sqrt{(\frac{(a-1)^2}{q})(\frac{(a+q)^2}{q})}$$$$\Rightarrow |a-p||b-q|=\sqrt{\frac{(a-1)^2(a+q)^2}{q^2}}$$$$\Rightarrow |a-p||b-q|=\sqrt{\frac{(a-1)(a+q)}{q}}^2$$$$\Rightarrow |a-p||b-q|=\sqrt{\frac{a^2+aq-a-q}{q}}^2$$$$\Rightarrow |a-p||b-q|=\sqrt{a\overline{a}-1}^2$$$$\Rightarrow |a-p||b-q| \text{ does not depend on } q$$$$\Rightarrow \boxed{AP\times BQ\text{ remains constant as }P \text{ varies}}_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$