如图，连接$EB$，$ED$.

$\because \angle ABC=\angle ADC=90^{\circ}$，$AE=EC$，

$\therefore EB=EA=EC=ED=\frac{1}{2}AC=\frac{1}{2}\times 12=6\left(cm\right)$，

$\because BF=FD$，

$\therefore EF\bot BD$，

$\because EA=EB$，

$\therefore \angle EAB=\angle BEA=15^{\circ}$，

$\therefore \angle BEC=\angle EAB+\angle EBA=30^{\circ}$，

$\because PB=PE$，

$\therefore \angle EBF=\angle BEC={30}^{\circ }$，

$\therefore EF=\frac{1}{2}BE=\frac{1}{2}\times 6=3\left(cm\right)$.

如图，连接$EB$，$ED$.

$\because \angle ABC=\angle ADC=90^{\circ}$，$AE=EC$，

$\therefore EB=EA=EC=ED=\frac{1}{2}AC=\frac{1}{2}\times 12=6\left(cm\right)$，

$\because BF=FD$，

$\therefore EF\bot BD$，

$\because EA=EB$，

$\therefore \angle EAB=\angle BEA=15^{\circ}$，

$\therefore \angle BEC=\angle EAB+\angle EBA=30^{\circ}$，

$\because PB=PE$，

$\therefore \angle EBF=\angle BEC={30}^{\circ }$，

$\therefore EF=\frac{1}{2}BE=\frac{1}{2}\times 6=3\left(cm\right)$.