20. (12 points) As shown in Figure 4, given that $PD$ is perpendicular to the plane of trapezoid $ABCD$, $\angle ADC = \angle BAD = 90^\circ$, $F$ is the midpoint of $PA$, $PD = \sqrt{2}$, $AB = AD = \frac{1}{2} CD = 1$. If quadrilateral $PDCE$ is a rectangle, and line segment $PC$ intersects $DE$ at point $N$.
(1) Prove: $AC \parallel$ plane $DEF$.
(2) Find the size of the dihedral angle $A-BC-P$.
(3) Does there exist a point $Q$ on line segment $EF$ such that the angle between $BQ$ and plane $BCP$ is $\frac{\pi}{6}$? If it exists, find the length of $FQ$; if not, explain the reason.
20. (1) Connect $F N$.

In $\triangle P A C$, since $F$ and $N$ are the midpoints of $P A$ and $P C$ respectively, we know $F N / / A C$.

Since $F N \subset$ plane $D E F$ and $A C \not \subset$ plane $D E F$, it follows that $A C / /$ plane $D E F$.
(2) As shown in Figure 6, take $D$ as the origin, and the lines $D A$, $D C$, and $D P$ as the $x$-axis, $y$-axis, and $z$-axis respectively, to establish a spatial rectangular coordinate system $D-x y z$.
Then $P(0,0, \sqrt{2}), B(1,1,0), C(0,2,0)$
$\Rightarrow \overrightarrow{P B}=(1,1,-\sqrt{2}), \overrightarrow{B C}=(-1,1,0)$.
Let the normal vector of plane $P B C$ be $\boldsymbol{m}=(x, y, z)$.
Then $\left\{\begin{array}{l}\boldsymbol{m} \cdot \overrightarrow{P B}=(x, y, z) \cdot(1,1,-\sqrt{2})=0, \\ \boldsymbol{m} \cdot \overrightarrow{B C}=(x, y, z) \cdot(-1,1,0)=0 .\end{array}\right.$
Let $x=1$, we get $\boldsymbol{m}=(1,1, \sqrt{2})$.
The normal vector of plane $A B C$ is $\boldsymbol{n}=(0,0,1)$, so $\cos \langle\boldsymbol{n}, \boldsymbol{m}\rangle=\frac{\boldsymbol{n} \cdot \boldsymbol{m}}{|\boldsymbol{n}||\boldsymbol{m}|}=\frac{\sqrt{2}}{2}$.
From Figure 6, we know that the dihedral angle $A-B C-P$ is an acute dihedral angle, hence the size of the dihedral angle $A-B C-P$ is $\frac{\pi}{4}$.
(3) Suppose there exists a point $Q$ that satisfies the condition.
Given $F\left(\frac{1}{2}, 0, \frac{\sqrt{2}}{2}\right), E(0,2, \sqrt{2})$, let $\overrightarrow{F Q}=\lambda \overrightarrow{F E}(0 \leqslant \lambda \leqslant 1)$.
Simplifying, we get $Q\left(\frac{1-\lambda}{2}, 2 \lambda, \frac{\sqrt{2}(1+\lambda)}{2}\right)$,
$\overrightarrow{B Q}=\left(-\frac{1+\lambda}{2}, 2 \lambda-1, \frac{\sqrt{2}(1+\lambda)}{2}\right)$.
Since the angle between line $B Q$ and plane $B C P$ is $\frac{\pi}{6}$, we have,
$$
\begin{array}{l}
\sin \frac{\pi}{6}=|\cos \langle\overrightarrow{B Q}, \boldsymbol{m}\rangle|=\left|\frac{\overrightarrow{B Q} \cdot \boldsymbol{m}}{|\overrightarrow{B Q}||\boldsymbol{m}|}\right| \\
=\frac{|5 \lambda-1|}{2 \sqrt{19 \lambda^{2}-10 \lambda+7}}=\frac{1}{2} \\
\Rightarrow \lambda^{2}=1 .
\end{array}
$$

Given $0 \leqslant \lambda \leqslant 1$, we know $\lambda=1$, meaning point $Q$ coincides with $E$. Therefore, there exists a point $Q$ on line segment $E F$, and
$|F Q|=|E F|=\frac{\sqrt{19}}{2}$.