Let's make an inversion in I with radius 1. $A'I$ is perpendicular to $B'C'$, because line $AI$ passes through a center of circle $BCI$. Let's denote centers of $BIP$ and $BIC$ by $O$ and $M$, respectively. Circle $BIP$ is tangent to line $AI$ at point $I$, so line $OI$ is perpendicular to $IM$. Thus $B'C' \perp B'P'$. So $A'I$ is parallel to $P'Q'$. Now we use absolutely fantastic formula for distance between points after inversion to rewrite thesis. And using some angle chasing and fact that circles $IA'B'$, $IB'C'$ and $IC'A'$ have equal radii we get: $A'R' = B'I$, $A'I = B'Q'$, $P'I = Q'I$ and $R'I = Q'I$. And it is sufficient to prove the new thesis.

Firstly, because $BI$ bisects $\angle{ABC} \equiv \angle{PBQ}$, we conclude that $I$ is the midpoint of arc $PQ$ not containing $B$, so $PI = IQ$. Because $AI$ is tangent to the circle at the midpoint of the arc $PQ$ not containing $B$, we conclude that $PQ$ is parallel to $AI$, so $\angle{PIA} = \angle{IPQ} = \angle{IQP} = \angle{(AI,IQ)} = \angle{AIR} \longrightarrow AI \text{ bisects } \angle{PIR}$. Hence, because $\angle{IAP} = \angle{IAR}$ and $\angle{PIA} = \angle{RIA}$, we get that $\triangle{API} \cong \triangle{ARI}$ in $A.A.S$ property. So $IP = IQ = IR$. Now, because $\angle{IAR} = \angle{IAP} = \angle{BPQ} = \angle{BIQ}$, and $\angle{AIR} = \angle{IPQ} = \angle{IBQ}$, then $\triangle{AIR} \sim \triangle{IBQ}$, thus $\frac{AR}{IQ} = \frac{IR}{BQ} \longrightarrow AR \cdot BQ = IQ \cdot IR = PI^2$.