答：$PC=PD$.

证明：过$P$分别作$PE\bot OB$于$E$，$PF\bot OA$于$F$，

$\therefore \angle CFP=\angle DEP=90^{\circ}$，

$\because OM$是$\angle AOB$的平分线，

$\therefore PE=PF$，

$\because \angle 1+\angle FPD=90^{\circ}$，$\angle AOB=90^{\circ}$，

$\therefore \angle FPE=90^{\circ}$，

$\therefore \angle 2+\angle FPD=90^{\circ}$，

$\therefore \angle 1=\angle 2$，

在$\triangle CFP$和$\triangle DEP$中，

$\left\{\begin{array}{}\angle CFP=\angle DEP \\ PE=PF \\ \angle 1=\angle 2\end{array}\right.$，

$\therefore \triangle CFP$≌$\triangle DEP\left(ASA\right)$，

$\therefore PC=PD$.

答：$PC=PD$.

证明：过$P$分别作$PE\bot OB$于$E$，$PF\bot OA$于$F$，

$\therefore \angle CFP=\angle DEP=90^{\circ}$，

$\because OM$是$\angle AOB$的平分线，

$\therefore PE=PF$，

$\because \angle 1+\angle FPD=90^{\circ}$，$\angle AOB=90^{\circ}$，

$\therefore \angle FPE=90^{\circ}$，

$\therefore \angle 2+\angle FPD=90^{\circ}$，

$\therefore \angle 1=\angle 2$，

在$\triangle CFP$和$\triangle DEP$中，

$\left\{\begin{array}{}\angle CFP=\angle DEP \\ PE=PF \\ \angle 1=\angle 2\end{array}\right.$，

$\therefore \triangle CFP$≌$\triangle DEP\left(ASA\right)$，

$\therefore PC=PD$.