Let $M=BL\cap(ABC)$, 1- With angle working you get that $M$ is the center of $(APC)$. 2- Let $K$ be the midpint of arc $APC$ in $(APC)$. With angle working you get $\angle PMB=\angle BMK$, then $K$ is the reflection of $P$ through $BM$. 3- Using power of point, $PL\cdot LQ=AL\cdot LC=BL\cdot LM$, then $BPMQ$ is cyclic, and $PM=MQ$ implies $\angle PBM=\angle MBQ$, then $B,K, Q$ are collinear and we are done.

hectorraul wrote: Let $M=BL\cap(ABC)$, 1- With angle working you get that $M$ is the center of $(APC)$. 2- Let $K$ be the midpint of arc $APC$ in $(APC)$. With angle working you get $\angle PMB=\angle BMK$, then $K$ is the reflection of $P$ through $BM$. 3- Using power of point, $PL\cdot LQ=AL\cdot LC=BL\cdot LM$, then $BPMQ$ is cyclic, and $PM=MQ$ implies $\angle PBM=\angle MBQ$, then $B,K, Q$ are collinear and we are done. Similar solution. A nicer approach than mine I would say.

Tuymaada 2012 #3