II. (This question is worth 25 points) Given the parabola $y^{2}=2 p x(p>0)$, with its focus at $F$. Question: Does there exist a chord $AB$ passing through $F$ (where $A$ and $B$ are both on the parabola, and $A$ is in the first quadrant, and a point $P$ on the positive $y$-axis), such that $P, A, B$ form a right triangle with $P$ as the right-angle vertex?
II. Restate the problem, and the equation of the straight line $AB$ is $y=\sqrt[4]{\frac{4}{3}}\left(x-\frac{p}{2}\right)$, and the y-coordinate of point $P$ is $p \sqrt[4]{\frac{3}{4}}\left(\frac{3}{2}+\right.$ $\frac{\sqrt{3}}{2}$ ). The proof process is given below.

As shown in the figure, the focus $F\left(\frac{p}{2}, 0\right), P\left(0, y_{P}\right), A\left(\frac{y_{A}^{2}}{2 p}, y_{A}\right), B$ $\left(\frac{y_{A}^{2}}{2 p}, y_{B}\right)$, where $y_{P}>0, y_{A} \neq y_{B}$. Let the slope of line $AB$ be $k$, then the equation of $AB$ is $y=k\left(x-\frac{p}{2}\right)$. Substituting it into the parabola equation gives $y=k\left(\frac{y^{2}}{2 p}-\frac{p}{2}\right)$. Therefore, $y_{A}$ and $y_{B}$ are the roots of the equation $k y^{2}-$ $2 p y-k p^{2}=0$. By the relationship between roots and coefficients, we have
$$
y_{A}+y_{B}=\frac{2 p}{k}, y_{A} \cdot y_{B}=-p^{2} .
$$

From $P A \perp P B$ and $P A=P B$, we get
$$
\left\{\begin{array}{l}
\frac{y_{A}-y_{P}}{\frac{y_{A}^{2}}{2 p}} \cdot \frac{y_{B}-y_{P}}{\frac{y_{B}^{2}}{2 p}}=-1, \\
\left(y_{A}-y_{P}\right)^{2}+\left(\frac{y_{A}^{2}}{2 p}\right)^{2}=\left(y_{B}-y_{P}\right)^{2}+\left(\frac{y_{B}^{2}}{2 p}\right)^{2} .
\end{array}\right.
$$

That is,
$$
\left\{\begin{array}{l}
y_{A} y_{B}-\left(y_{A}+y_{B}\right) y_{P}+y_{P}^{2}+\frac{\left(y_{A} y_{B}\right)^{2}}{4 p^{2}}=0, \\
\left(y_{A}-y_{B}\right)\left[\left(y_{A}+y_{B}-2 y_{P}\right)\right. \\
\left.+\frac{\left(y_{A}+y_{B}\right)\left(\left(y_{A}+y_{B}\right)^{2}-2 y_{A} y_{B}\right)}{4 p^{2}}\right]=0 .
\end{array}\right.
$$
Substituting (1) into the above two equations and canceling $y_{A}-y_{B}$ in the second equation, we get
$$
\left\{\begin{array}{l}
-p^{2}-\frac{2 p}{k} y_{p}+y_{p}^{2}+\frac{p^{4}}{4 p^{2}}=0, \\
\frac{2 p}{k}-2 y_{p}+\frac{\frac{2 p}{k}\left(\frac{4 p^{2}}{k^{2}}+2 \cdot p^{2}\right)}{4 p^{2}}=0 .
\end{array}\right.
$$

Solving for $y_{p}$ in the second equation, we get $y_{p}=\frac{p}{k}\left(\frac{3}{2}+\frac{1}{k^{2}}\right)$, and substituting it into the first equation, we get
$$
3 k^{6}+3 k^{4}-4 k^{2}-4=0,
$$

which is $\left(3 k^{4}-4\right)\left(k^{2}+1\right)=0$.
Thus, we get $k=\sqrt[4]{\frac{4}{3}}$ (discarding the negative value),
$$
y_{\mathrm{P}}=p \sqrt[4]{\frac{3}{4}}\left(\frac{3+\sqrt{3}}{2}\right) \text {. }
$$
$y_{A}$ and $y_{B}$ can both be found, so the chord $AB$ that meets the problem's requirements definitely exists and is uniquely determined.