9. (14 points) As shown in Figure 1, given the parabola $y^{2} = 2 p x (p > 0)$, with its focus at $F$, a line passing through the focus $F$ and inclined at an angle $\theta$ $(0 < \theta < \pi)$ intersects the parabola at points $A$ and $B$. Connecting $A O$ (where $O$ is the origin), it intersects the directrix at point $B^{\prime}$. Connecting $B O$, it intersects the directrix at point $A^{\prime}$. Find the area of quadrilateral $A B B^{\prime} A^{\prime}$.
When $\theta=\frac{\pi}{2}$, $s_{\text {quadrilateral } B B B K^{\prime}}=2 p^{2}$.
When $\theta \neq \frac{\pi}{2}$, let $k=\tan \theta$.
Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$.
From $y=k\left(x-\frac{p}{2}\right)$ and $y^{2}=2 p x$, eliminating $x$ gives
$$
y^{2}-\frac{2 p}{k} y-p^{2}=0 \text {. }
$$

Then $y_{1}+y_{2}=\frac{2 p}{k}$ and $y_{1} y_{2}=-p^{2}$.
The equation of line $A O$ is $y=\frac{y_{1}}{x_{1}} x$, which is $y=\frac{2 p}{y_{1}} x$. Therefore, the coordinates of the intersection point of $A O$ with the directrix are $B^{\prime}\left(-\frac{p}{2},-\frac{p^{2}}{y_{1}}\right)$. From equation (1), we know $y_{2}=-\frac{p^{2}}{y_{1}}$, so the y-coordinates of points $B$ and $B^{\prime}$ are equal. Thus, $B B^{\prime} / / x$ axis.
Similarly, $A A^{\prime} / / x$ axis.
Therefore, quadrilateral $A B B^{\prime} A^{\prime}$ is a right trapezoid, and its area is
$S_{\text {quadrilateral } A B B A^{\prime}}=\frac{1}{2}\left(\left|A A^{\prime}\right|+\left|B B^{\prime}\right|\right) \cdot\left|A^{\prime} B^{\prime}\right|$ $=\frac{1}{2}|A B| \cdot\left|A^{\prime} B^{\prime}\right|$
$=\frac{1}{2} \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \cdot\left|y_{2}-y_{1}\right|$
$=\frac{1}{2}\left(y_{2}-y_{1}\right)^{2} \sqrt{1+\frac{1}{k^{2}}}$
$=\frac{1}{2} \sqrt{1+\frac{1}{k^{2}}}\left[\left(y_{1}+y_{2}\right)^{2}-4 y_{1} y_{2}\right]$
$=2 p^{2}\left(1+\frac{1}{k^{2}}\right)^{\frac{3}{2}}=2 p^{2}\left(1+\cot ^{2} \theta\right)^{\frac{3}{2}}$.