C

设正方形$ABCD$的边长为$x$，则$AE=\frac{1}{2}x$，
$\because$四边形$ABCD$是正方形，
$\therefore \angle DAB=\angle ABC=90^{\circ}$，$AD$∥$BC$，
在$Rt\triangle ABE$中，由勾股定理得：$BE=\sqrt{A{B}^{2}+A{E}^{2}}=\sqrt{{x}^{2}+(\frac{1}{2}x)^{2}}=\frac{\sqrt{5}}{2}x=EF$，
$\therefore AF=EF-AE=\frac{\sqrt{5}}{2}x-\frac{1}{2}x=\frac{\sqrt{5}-1}{2}x$，
$\therefore$正方形$AFGH$的边长为$\frac{\sqrt{5}-1}{2}x$，
$\therefore BH=AB-AH=x-\frac{\sqrt{5}-1}{2}x=\frac{3-\sqrt{5}}{2}x$，
$\because FH$为正方形$AFGH$的对角线，
$\therefore \angle FHA=\angle BHQ=45^{\circ}$，
$\therefore \triangle BHQ$是等腰直角三角形，
$\therefore BH=BQ=\frac{3-\sqrt{5}}{2}x$，
$\because EF$∥$BQ$，
$\therefore \triangle EFP$∽$\triangle BQP$，
$\therefore \frac{{S}_{△EFP}}{{S}_{△BQP}}=(\frac{EF}{BQ})^{2}=(\frac{\frac{\sqrt{5}}{2}}{\frac{3-\sqrt{5}}{2}})^{2}=\frac{5}{14-6\sqrt{5}}$，
$\therefore S_{\triangle EFP}-S_{\triangle BPQ}=(\frac{5}{14-6\sqrt{5}}-1)S_{\triangle BPQ}=6\sqrt{5}-9$，
解得：$S_{\triangle BPQ}=14-6\sqrt{5}$，
设$\triangle EFP$在$EF$边上的高为$h_{1}$，$\triangle BPQ$在$BQ$边上的高为$h_{2}$，
则$\frac{{h}_{1}}{{h}_{2}}=\frac{EF}{BQ}=\frac{\frac{\sqrt{5}}{2}}{\frac{3-\sqrt{5}}{2}}=\frac{\sqrt{5}}{3-\sqrt{5}}$，
$\therefore h_{1}+h_{2}=(\frac{\sqrt{5}}{3-\sqrt{5}}+1)h_{2}$，
$\because h_{1}+h_{2}=AB=x$，
$\therefore (\frac{\sqrt{5}}{3-\sqrt{5}}+1)h_{2}=x$，
$\therefore h_{2}=\frac{3-\sqrt{5}}{3}x$，
$\therefore S_{\triangle BPQ}=\frac{1}{2}BQ\cdot h_{2}=\frac{1}{2}\times \frac{3-\sqrt{5}}{2}x\times \frac{3-\sqrt{5}}{3}x=14-6\sqrt{5}$，
解得：$x=2\sqrt{3}$，
$\therefore AE=\frac{1}{2}x=\frac{1}{2}\times 2\sqrt{3}=\sqrt{3}$，
故选：$C$.