如图，在四棱锥${{P}-{ABCD}}$中，$AD//BC$，$AB\perp BC$，$AB=BC=1$，$AD=a\left( a\gt 1 \right)$，$PA\perp $平面$ABCD$，${ {PD}}$与平面$ABCD$所成角为$45{}^\circ $，$E$为${ {PD}}$中点，

$(1)$证明：$BE\perp PD$；

$(2)$若直线$PC$与平面$ABE$所成角为$60{}^\circ $，求$a$的值$.$

$(1)$证明见解析；

$(2)$$a=2$$.$

$(1)$$\because AD//BC$，$AB\perp BC$，

$\therefore AD\perp AB$，

$\because PA\perp $平面$ABCD$，${ {PD}}$与平面$ABCD$所成角为$45{}^\circ $，

$\therefore \angle PDA$为${ {PD}}$与平面$ABCD$所成角，即$\angle PDA=45{}^\circ $，

则$PD=AD=a\left( a\gt 1 \right)$，又$AD,AB\subset $平面$ABCD$，

$\therefore PA\perp AD,PA\perp AB$，

$\therefore $ 可建立如图所示的空间直角坐标系$A-xyz$，

则由题$A\left( 0,0,0 \right),B\left( 1,0,0 \right),E\left( 0,\dfrac{a}{2},\dfrac{a}{2} \right),P\left( 0,0,a \right),D\left( 0,a,0 \right),C\left( 1,1,0 \right)$，

$\therefore \overrightarrow{BE}=\left( -1,\dfrac{a}{2},\dfrac{a}{2} \right),\overrightarrow{PD}=\left( 0,a,-a \right),\overrightarrow{PC}=\left( 1,1,-a \right)$，$\overrightarrow{AB}=\left( 1,0,0 \right)$，

$\therefore \overrightarrow{BE}\cdot \overrightarrow{PD}=\left( -1,\dfrac{a}{2},\dfrac{a}{2} \right)\cdot \left( 0,a,-a \right)=\dfrac{{{a}^{2}}}{2}-\dfrac{{{a}^{2}}}{2}=0$，

$\therefore \overrightarrow{BE}\perp \overrightarrow{PD}$，即$BE\perp PD$$.$

$(2)$设平面$ABE$的法向量为$\boldsymbol{n}=\left( x,y,z \right)$，

则$\begin{cases} \boldsymbol{n}\perp \overrightarrow{BE} \\ \boldsymbol{n}\perp \overrightarrow{AB} \\ \end{cases}$，

$\therefore \begin{cases} \boldsymbol{n}\cdot \overrightarrow{BE}=0 \\ \boldsymbol{n}\cdot \overrightarrow{AB}=0 \\ \end{cases}$，

$\therefore $ 由$(1)$得$\begin{cases} -x+\dfrac{a}{2}y+\dfrac{a}{2}z=0 \\ x=0 \\ \end{cases}$，

取$y=-1$，则$\boldsymbol{n}=\left( 0,-1,1 \right)$，又直线$PC$与平面$ABE$所成角为$60{}^\circ $，

$\therefore \sin 60{}^\circ =\left| \cos\langle \boldsymbol{n},\overrightarrow{PC} \rangle\right|=\left| \dfrac{\boldsymbol{n}\cdot \overrightarrow{PC}}{\left| {\boldsymbol{n}} \right|\left| \overrightarrow{PC} \right|} \right|=\dfrac{a+1}{\sqrt{{{1}^{2}}+{{1}^{2}}+{{\left( -a \right)}^{2}}}\sqrt{{{\left( -1 \right)}^{2}}+{{1}^{2}}}}$

$=\dfrac{a+1}{\sqrt{2{{a}^{2}}+4}}=\dfrac{\sqrt{3}}{2}$，解得$a=2$$.$