Let's see that $Q$ is the orthocenter of $\triangle PAB$, Indeed notice that $BP$ bisects $\angle BQC$, and also since $BD=DC$ we have $\angle CDB=\angle CAB=\angle CBD$, and then $\angle BQC=180^{\circ}-\angle QCB-\angle QBC=180^{\circ}-\angle QCB-\angle CAB= \angle CBA=\angle BCA$, so $\triangle BQC$ is isosceles and $BP$ bisects $\angle BQC$ so $BP\perp QC$, in a similar way $AP\perp QD$, so $Q$ is the orthocenter of $\triangle PAB$.

Ricochet wrote: Let's see that $Q$ is the orthocenter of $\triangle PAB$, Indeed notice that $BP$ bisects $\angle BQC$, and also since $BD=DC$ we have $\angle CDB=\angle CAB=\angle CBD$, and then $\angle BQC=180^{\circ}-\angle QCB-\angle QBC=180^{\circ}-\angle QCB-\angle CAB= \angle CBA=\angle BCA$, so $\triangle BQC$ is isosceles and $BP$ bisects $\angle BQC$ so $BP\perp QC$, in a similar way $AP\perp QD$, so $Q$ is the orthocenter of $\triangle PAB$. Information at problem is $BC=CD$ But its okay since still get $\angle{CDB}=\angle{CBD}$