延长$AP$交$BC$于点$D$，如图所示：

$\because BP$平分$\angle ABC$，
$\therefore \angle ABP=\angle DBP$，
$\because AP\bot BP$，
$\therefore \angle APB=\angle DPB=90^{\circ}$，
在$\triangle ABP$和$\triangle DBP$中，
$\left\{\begin{array}{l}{∠ABP=∠DBP}\\{BP=BP}\\{∠APB=∠DPB=90°}\end{array}\right.$，
$\therefore \triangle ABP$≌$\triangle DBP\left(ASA\right)$，
$\therefore S_{\triangle ABP}=S_{\triangle DBP}=5$，$AP=DP$，
$\therefore S_{\triangle ABD}=S_{\triangle ABP}+S_{\triangle DBP}=10$，
又$\because S_{\triangle ABC}=14$，
$\therefore S_{\triangle ACD}=S_{\triangle ABC}-S_{\triangle ABD}=4$，
$\because AP=DP$，
$\therefore S_{\triangle APC}=S_{\triangle DPC}=\frac{1}{2}S_{\triangle ACD}=2$.
故答案为：$2$.

延长$AP$交$BC$于点$D$，如图所示：

$\because BP$平分$\angle ABC$，
$\therefore \angle ABP=\angle DBP$，
$\because AP\bot BP$，
$\therefore \angle APB=\angle DPB=90^{\circ}$，
在$\triangle ABP$和$\triangle DBP$中，
$\left\{\begin{array}{l}{∠ABP=∠DBP}\\{BP=BP}\\{∠APB=∠DPB=90°}\end{array}\right.$，
$\therefore \triangle ABP$≌$\triangle DBP\left(ASA\right)$，
$\therefore S_{\triangle ABP}=S_{\triangle DBP}=5$，$AP=DP$，
$\therefore S_{\triangle ABD}=S_{\triangle ABP}+S_{\triangle DBP}=10$，
又$\because S_{\triangle ABC}=14$，
$\therefore S_{\triangle ACD}=S_{\triangle ABC}-S_{\triangle ABD}=4$，
$\because AP=DP$，
$\therefore S_{\triangle APC}=S_{\triangle DPC}=\frac{1}{2}S_{\triangle ACD}=2$.
故答案为：$2$.