延长$BA$，作$PN\bot BD$，$PF\bot BA$，$PM\bot AC$，

设$\angle PCD=x^{\circ}$，

$\because CP$平分$\angle ACD$，

$\therefore \angle ACP=\angle PCD=x^{\circ}$，$PM=PN$，

$\because BP$平分$\angle ABC$，

$\therefore \angle ABP=\angle PBC$，$PF=PN$，

$\therefore PF=PM$，

$\because \angle BPC=40^{\circ}$，

$\therefore \angle ABP=\angle PBC=\angle PCD-\angle BPC=\left(x-40\right)^{\circ}$，

$\therefore \angle BAC=\angle ACD-\angle ABC=2x^{\circ}-\left(x^{\circ}-40^{\circ}\right)-\left(x^{\circ}-40^{\circ}\right)=80^{\circ}$，

$\therefore \angle CAF=100^{\circ}$，

在$Rt\triangle PFA$和$Rt\triangle PMA$中，

$\because \left\{\begin{array}{}PA=PA \\ PM=PF\end{array}\right.$，

$\therefore Rt\triangle PFA$≌$Rt\triangle PMA\left(HL\right)$，

$\therefore \angle FAP=\angle PAC=50^{\circ}$.

故答案为：$50^{\circ}$.

延长$BA$，作$PN\bot BD$，$PF\bot BA$，$PM\bot AC$，

设$\angle PCD=x^{\circ}$，

$\because CP$平分$\angle ACD$，

$\therefore \angle ACP=\angle PCD=x^{\circ}$，$PM=PN$，

$\because BP$平分$\angle ABC$，

$\therefore \angle ABP=\angle PBC$，$PF=PN$，

$\therefore PF=PM$，

$\because \angle BPC=40^{\circ}$，

$\therefore \angle ABP=\angle PBC=\angle PCD-\angle BPC=\left(x-40\right)^{\circ}$，

$\therefore \angle BAC=\angle ACD-\angle ABC=2x^{\circ}-\left(x^{\circ}-40^{\circ}\right)-\left(x^{\circ}-40^{\circ}\right)=80^{\circ}$，

$\therefore \angle CAF=100^{\circ}$，

在$Rt\triangle PFA$和$Rt\triangle PMA$中，

$\because \left\{\begin{array}{}PA=PA \\ PM=PF\end{array}\right.$，

$\therefore Rt\triangle PFA$≌$Rt\triangle PMA\left(HL\right)$，

$\therefore \angle FAP=\angle PAC=50^{\circ}$.

故答案为：$50^{\circ}$.