证明略，直线方程为 $x=-1$

设 $A(x_{1},y_{1})$，$B(x_{2},y_{2})$，则 $l_{1}$ 方程为$$y_{1}y=2(x+x_{1}),$$直线 $l_{2}$ 方程为$$y_{2}y=2(x+x_{2}).$$设 $P(x_{0},y_{0})$，则$$\begin{split}y_{1}y_{0}=2(x_{0}+x_{1}),\\y_{2}y_{0}=2(x_{0}+x_{2}),\end{split}$$所以点 $A(x_{1},y_{1})$，$B(x_{2},y_{2})$ 坐标满足方程$$yy_{0}=2(x_{0}+x),$$故直线 $AB$ 方程为\[yy_{0}=2(x_{0}+1),\]因为直线 $AB$ 过 $F(1,0)$，因此 $x_{0}=-1$，点 $P$ 在定直线 $x=-1$ 上．

$\dfrac{128\sqrt 3}{9}$

由第 $(1)$ 小题知，$C,D$ 的坐标分别为$$C\left(4,\dfrac{8}{y_{1}}+\dfrac{1}{2}y_{1}\right) , D\left(4,\dfrac{8}{y_{2}}+\dfrac{1}{2}y_{2}\right),$$所以\[\begin{split}|CD|&=\left|\left(\dfrac{8}{y_{1}}+\dfrac{1}{2}y_{1}\right)-\left(\dfrac{8}{y_{2}}+\dfrac{1}{2}y_{2}\right)\right|\\ &=\left|\dfrac{(y_{1}y_{2}-16)(y_{1}-y_{2})}{2y_{1}y_{2}}\right|,\end{split}\]故\[S_{\triangle PCD}=\dfrac{1}{2}\left|4-\dfrac{y_{1}y_{2}}{4}\right|\cdot \left|\dfrac{(y_{1}y_{2}-16)(y_{1}-y_{2})}{2y_{1}y_{2}}\right|.\]设 $y_{1}y_{2}=-t^{2}(t>0)$，$|y_{1}-y_{2}|=m$，由\[(y_{1}+y_{2})^{2}=(y_{1}-y_{2})^{2}+4y_{1}y_{2}=m^{2}-4t^{2}\geqslant 0,\]可知 $m\geqslant 2t$，当且仅当 $y_{1}+y_{2}=0$ 时等号成立．因此\[\begin{split}S_{\triangle PCD}&=\dfrac{1}{2}\left|4+\dfrac{t^{2}}{4}\right|\cdot \left|\dfrac{(-t^{2}-16)m}{-2t^{2}}\right|\\&=\dfrac{m\cdot (t^{2}+16)^{2}}{16t^{2}}\\ &\geqslant \dfrac{2t\cdot (t^{2}+16)^{2}}{16t^{2}}\\&=\dfrac{(t^{2}+16)^{2}}{8t}.\end{split}\]设 $f(t)=\dfrac{(t^{2}+16)^{2}}{8t}$，则\[\begin{split}f'(t)&=\dfrac{2(t^{2}+16)\cdot 2t\cdot t-(t^{2}+16)^{2}}{8t^{2}}\\ &=\dfrac{(3t^{2}-16)(t^{2}+16)}{8t^{2}},\end{split}\]所以当 $0<t<\dfrac{4\sqrt 3}{3}$ 时，$f'(t)<0$；当 $t>\dfrac{4\sqrt 3}{3}$ 时，$f'(t)>0$．
因此 $f(t)$ 在区间 $\left(0,\dfrac{4\sqrt 3}{3}\right]$ 上为减函数；在区间 $\left[\dfrac{4\sqrt 3}{3},+\infty\right)$ 上为增函数，故当 $t=\dfrac{4\sqrt 3}{3}$ 时，$f(t)$ 取最小值 $\dfrac{128\sqrt 3}{9}$．
因此当 $y_{1}+y_{2}=0$，$y_{1}y_{2}=-\dfrac{16}{3}$，即 $y_{1}=\dfrac{4}{\sqrt 3}$，$y_{2}=-\dfrac{4}{\sqrt 3}$ 时，$\triangle PCD$ 面积取最小值 $\dfrac{128\sqrt 3}{9}$．