如图，在四棱锥${{P}-{ABCD}}$中，平面$ABCD\perp $平面$PAD$，$AD//BC$，$AB=BC=PA=1$，$AD=2$，$\angle ADP=30{}^\circ $，$\angle BAD=90{}^\circ $，$E$是$PD$的中点．

$(1)$求证：$PD\perp PB$；

$(2)$若点$M$在线段$PC$上，异面直线$BM$和$CE$所成角的余弦值为$\dfrac{\sqrt{10}}{5}$，求面$MAB$与面$PCD$夹角的余弦值．

$(1)$证明见解析；

$(2)$$\\dfrac{2\\sqrt{105}}{35}$

$(1)$在$\triangle PAD$中，

$∵$$PA=1$，$AD=2$，$\angle ADP=30{}^\circ $，

由余弦定理可得：$P{{A}^{2}}=A{{D}^{2}}+P{{D}^{2}}-2AD\cdot PD\cdot \cos \angle ADP$，

即$1=4+P{{D}^{2}}-4PD\cdot \cos {{30}^{\text{o}}}\Rightarrow PD=\sqrt{3}$，

$∴$ $A{{D}^{2}}=P{{A}^{2}}+P{{D}^{2}}$，

从而$PD\perp PA$

$∵$$\angle BAD=90{}^\circ $，

$∴$$AB\perp AD$

$∵$ 平面$ABCD\perp $平面$PAD$，平面$ABCD$$\text{}$平面$PAD$$=AD$，$AB$$\subset $平面$ABC$  $D$.

$∴$ $AB\perp $平面$PAD$，

$∴$ $P D \subset$平面$PAD$，

$∴$ $PD\perp AB$$.$

$∵$ $AB\cap PA=A$，$AB$$\subset $平面$PAB$，$PA$$\subset $平面$PAB$，

$∴$ $PD\perp $平面$PA$  $B$.

$∵$ $PB\subset $平面$PAB$，

$∴$ $PD\perp PB$．

$(2)$以$A$为原点，以$AD$为$y$轴，建系如图所示，

则$B\left( 0,0,1 \right)$，$P\left( \dfrac{\sqrt{3}}{2},\dfrac{1}{2},0 \right)$，$C\left( 0,1,1 \right),D\left( 0,2,0 \right)$，$E\left( \dfrac{\sqrt{3}}{4},\dfrac{5}{4},0 \right)$，

则$\overrightarrow{CE}=\left( \dfrac{\sqrt{3}}{4},\dfrac{1}{4},-1 \right)$，$\overrightarrow{BP}=\left( \dfrac{\sqrt{3}}{2},\dfrac{1}{2},-1 \right)$，$\overrightarrow{PC}=\left( -\dfrac{\sqrt{3}}{2},\dfrac{1}{2},1 \right)$

$\overrightarrow{AB}=\left( 0,0,1 \right)$，$\overrightarrow{CD}=\left( 0,1,-1 \right)$$.$

设$\overrightarrow{PM}=\lambda\overrightarrow{PC}\left( 0\leqslant\lambda\leqslant1\right)$，则$\overrightarrow{BM}=\overrightarrow{BP}+\overrightarrow{PM}=\overrightarrow{BP}+\lambda \overrightarrow{PC}$

$=\left( \dfrac{\sqrt{3}}{2},\dfrac{1}{2},-1 \right)+\lambda \left( -\dfrac{\sqrt{3}}{2},\dfrac{1}{2},1 \right)=\left( \dfrac{\sqrt{3}}{2}\left( 1-\lambda \right),\dfrac{1}{2}\left( 1+\lambda \right),\lambda -1 \right)$

设异面直线$BM$和$CE$所成角为$\alpha $，则

$\cos \alpha =\left| \cos \left\langle \overrightarrow{BM},\overrightarrow{CE} \right\rangle \right|=\left| \dfrac{\overrightarrow{BM}\cdot \overrightarrow{CE}}{\left| \overrightarrow{BM} \right|\cdot \left| \overrightarrow{CE} \right|} \right|=\dfrac{\left| 6-5\lambda \right|}{2\sqrt{5}\times \sqrt{2{{\lambda }^{2}}-3\lambda +2}}=\dfrac{\sqrt{10}}{5}$

得$\lambda =\dfrac{2}{3}$$.$此时，$\overrightarrow{BM}=\left( \dfrac{\sqrt{3}}{6},\dfrac{5}{6},-\dfrac{1}{3} \right).$

设面$MAB$的一个法向量为$\boldsymbol{{n_1}}=\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$，

有$\begin{cases} \boldsymbol{{n_1}}\cdot \overrightarrow{AB}=0 \\ \boldsymbol{{n_1}}\cdot \overrightarrow{BM}=0 \\ \end{cases}\Rightarrow \begin{cases} {{z}_{1}}=0 \\ \dfrac{\sqrt{3}}{6}{{x}_{1}}+\dfrac{5}{6}{{y}_{1}}-\dfrac{1}{3}{{z}_{1}}=0 \\ \end{cases}$

令${{y}_{1}}=\sqrt{3}$，则${{x}_{1}}=-5$，${{z}_{1}}=0$，取 $\boldsymbol{{n_1}}=\left( -5,\sqrt{3},0 \right)$$.$

设面$PCD$的一个法向量为$\boldsymbol{{n_2}}=\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$，

有$\begin{cases} \boldsymbol{{n_2}}\cdot \overrightarrow{CD}=0 \\ \boldsymbol{{n_2}}\cdot \overrightarrow{PC}=0 \\ \end{cases}\Rightarrow \begin{cases} {{y}_{2}}-{{z}_{2}}=0 \\ -\dfrac{\sqrt{3}}{2}{{x}_{2}}+\dfrac{1}{2}{{y}_{2}}+{{z}_{2}}=0 \\ \end{cases}$

令${{x}_{2}}=\sqrt{3}$，则${{y}_{2}}=1$，$z_2=1$，取$\boldsymbol{{n_2}}=\left( \sqrt{3},1,1 \right)$

设面$MAB$与面$PCD$的夹角为$\theta $，

则$\cos \theta =\left| \dfrac{{\boldsymbol{n}_{1}}\cdot {\boldsymbol{n}_{2}}}{\left| {\boldsymbol{n}_{1}} \right|\cdot \left| {\boldsymbol{n}_{2}} \right|} \right|=\left| \dfrac{-4\sqrt{3}}{\sqrt{140}} \right|=\dfrac{2\sqrt{105}}{35}$$.$

即面$MAB$与面$PCD$夹角的余弦值为$\dfrac{2\sqrt{105}}{35}$$.$