$(1)$$\because$四边形$ABCD$为正方形，
$\therefore AD=AD$，$\angle BAD=90^{\circ}$，
$\because AE=AB$，
$\therefore AE=AD$，$\angle ABE=\angle AEB$，
$\therefore \angle ADE=\angle AED$，
$\because \angle BAD+\angle ABE+\angle BED+\angle ADE=360^{\circ}$，$\angle BED=\angle AEB+\angle AED$，
$\therefore 90^{\circ}+2\angle BED=360^{\circ}$，
$\therefore \angle BED=135^{\circ}$，
故$\angle BED$的度数为$135^{\circ}$.
$(2)$①证明：$\because CE=CF$，
$\therefore \angle CEF=\angle CFE$，
$\because$四边形$ABCD$为正方形，
$\therefore AB$∥$CD$，
$\therefore \angle CFE=\angle ABE$，
$\because \angle ABE=\angle AEB$，$\angle CEF=\angle CFE$，
$\therefore \angle CEF=\angle AEB$，
$\because \angle AEB+\angle AEF=180^{\circ}$，
$\therefore \angle CEF+\angle AEF=180^{\circ}$，
$\therefore A$，$E$，$C$三点共线，
②过点$A$作$AG\bot BF$于点$G$，则$BG=GE$，
$\because CE\bot BF$，
$\therefore \angle BEC=AGB=90^{\circ}$，$\angle CBE+\angle BCE=90^{\circ}$，
$\because$四边形$ABCD$为正方形，
$\therefore AB=BC$，$\angle ABC=90^{\circ}$，
$\therefore \angle ABG=\angle CBE=90^{\circ}$，
$\therefore \angle ABG=\angle BCE$，
$\therefore \triangle ABG$≌$\triangle BCE\left(ASA\right)$，
$\therefore BG=CE$，
$\therefore BE=2CE$，
$\because \angle BED=135^{\circ}$，
$\therefore \angle BEP=45^{\circ}$，
$\because \angle BEC=90^{\circ}$，
$\therefore \angle BEP=\angle CEP$，
过点$P$分别作$PM\bot BE$于点$M$，$PN\bot CN$于点$N$，则$PM=PN$，
$\because S_{\triangle BPE}$：$S_{\triangle CPE}=\frac{1}{2}BE\cdot PM$：$\frac{1}{2}CE•PN$，
$\therefore \frac{BP}{PC}=\frac{BE}{CE}=2$.
故$\frac{BP}{PC}$的长为$2$.

$(1)$$\because$四边形$ABCD$为正方形，
$\therefore AD=AD$，$\angle BAD=90^{\circ}$，
$\because AE=AB$，
$\therefore AE=AD$，$\angle ABE=\angle AEB$，
$\therefore \angle ADE=\angle AED$，
$\because \angle BAD+\angle ABE+\angle BED+\angle ADE=360^{\circ}$，$\angle BED=\angle AEB+\angle AED$，
$\therefore 90^{\circ}+2\angle BED=360^{\circ}$，
$\therefore \angle BED=135^{\circ}$，
故$\angle BED$的度数为$135^{\circ}$.
$(2)$①证明：$\because CE=CF$，
$\therefore \angle CEF=\angle CFE$，
$\because$四边形$ABCD$为正方形，
$\therefore AB$∥$CD$，
$\therefore \angle CFE=\angle ABE$，
$\because \angle ABE=\angle AEB$，$\angle CEF=\angle CFE$，
$\therefore \angle CEF=\angle AEB$，
$\because \angle AEB+\angle AEF=180^{\circ}$，
$\therefore \angle CEF+\angle AEF=180^{\circ}$，
$\therefore A$，$E$，$C$三点共线，
②过点$A$作$AG\bot BF$于点$G$，则$BG=GE$，
$\because CE\bot BF$，
$\therefore \angle BEC=AGB=90^{\circ}$，$\angle CBE+\angle BCE=90^{\circ}$，
$\because$四边形$ABCD$为正方形，
$\therefore AB=BC$，$\angle ABC=90^{\circ}$，
$\therefore \angle ABG=\angle CBE=90^{\circ}$，
$\therefore \angle ABG=\angle BCE$，
$\therefore \triangle ABG$≌$\triangle BCE\left(ASA\right)$，
$\therefore BG=CE$，
$\therefore BE=2CE$，
$\because \angle BED=135^{\circ}$，
$\therefore \angle BEP=45^{\circ}$，
$\because \angle BEC=90^{\circ}$，
$\therefore \angle BEP=\angle CEP$，
过点$P$分别作$PM\bot BE$于点$M$，$PN\bot CN$于点$N$，则$PM=PN$，
$\because S_{\triangle BPE}$：$S_{\triangle CPE}=\frac{1}{2}BE\cdot PM$：$\frac{1}{2}CE•PN$，
$\therefore \frac{BP}{PC}=\frac{BE}{CE}=2$.
故$\frac{BP}{PC}$的长为$2$.