证明略，直线方程为 $x=-1$

设 $A\left(\dfrac{y_{1}^{2}}{4},y_{1}\right)$，$B\left(\dfrac{y_{2}^{2}}{4},y_{2}\right)(y_{1}>0>y_{2})$．
易知 $l_{1}$ 斜率存在，设为 $k_{1}$，则 $l_{1}$ 方程为$$y-y_{1}=k\left(x-\dfrac{y_{1}^{2}}{4}\right).$$由$$\begin{cases}y-y_{1}=k_{1}\left(x-\dfrac{y_{1}}{4}\right),\\ y^{2}=4x,\end{cases}$$得\[k_{1}y^{2}-4y+4y_{1}-k_{1}y_{1}^{2}=0.\]由直线 $l_{1}$ 与抛物线 $C$ 相切，知\[\Delta =16-4k_{1}(4y_{1}-k_{1}y_{1}^{2})=0,\]解得 $k_{1}=\dfrac{2}{y_{1}}$，所以 $l_{1}$ 方程为$$y=\dfrac{2}{y_{1}}x+\dfrac{1}{y_{1}}.$$同理，$l_{2}$ 方程为$$y=\dfrac{2}{y_{2}}x+\dfrac{1}{2}y_{2}.$$联立 $l_{1},l_{2}$ 方程可得点 $P$ 坐标 $P\left(\dfrac{y_{1}y_{2}}{4},\dfrac{y_{1}+y_{2}}{2}\right)$．
因为\[k_{AB}=\dfrac{y_{1}-y_{2}}{\dfrac{y_{1}^{2}}{4}-\dfrac{y_{2}^{2}}{4}}=\dfrac{4}{y_{1}+y_{2}},\]所以 $AB$ 方程为$$y-y_{1}=\dfrac{4}{y_{1}+y_{2}}\left(x-\dfrac{y_{1}^{2}}{4}\right),$$又因为 $AB$ 过抛物线 $C$ 的焦点 $F(1,0)$，所以$$-y_{1}=\dfrac{4}{y_{1}+y_{2}}\left(1-\dfrac{y_{1}^{2}}{4}\right),$$解得 $y_{1}y_{2}=-4$．
因此$$x_{P}=\dfrac{y_{1}y_{2}}{4}=-1,$$故点 $P$ 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 $\dfrac{128\sqrt 3}{9}$．