Example 5 Given a quadrilateral pyramid $P-ABCD$, the base $ABCD$ is a rhombus with side length 2, $\angle ABC=60^{\circ}, E$ is the midpoint of $AB$, $PA \perp$ plane $ABCD$, and the sine of the angle formed by $PC$ and plane $PAD$ is $\frac{\sqrt{6}}{4}$.
(1) Find a point $F$ on edge $PD$ such that $AF \parallel$ plane $PEC$;
(2) Find the cosine of the dihedral angle $D-PE-A$. ${ }^{[5]}$
【Analysis】(1) Let the intersection of $A C$ and $B D$ be $O$. Establish a spatial rectangular coordinate system with $B D$ as the $x$-axis, $C A$ as the $y$-axis, and a line through point $O$ perpendicular to plane $A B C D$ as the $z$-axis, as shown in Figure 9.
Then $A(0,1,0), B(-\sqrt{3}, 0,0), C(0,-1,0)$,
$D(\sqrt{3}, 0,0), E\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}, 0\right)$.
Let $P(0,1, m)$. Then $\overrightarrow{P C}=(0,-2,-m)$.
Let the normal vector of plane $P A D$ be $\boldsymbol{n}=(x, y, z)$,
$\overrightarrow{A P}=(0,0, m), \overrightarrow{A D}=(\sqrt{3},-1,0)$.
Then $\left\{\begin{array}{l}m z=0, \\ \sqrt{3} x-y=0\end{array} \Rightarrow \boldsymbol{n}=(\sqrt{3}, 3,0)\right.$
$\Rightarrow|\cos \langle\overrightarrow{P C}, \boldsymbol{n}\rangle|=\left|\frac{-6}{\sqrt{4+m^{2}} \sqrt{12}}\right|=\frac{\sqrt{6}}{4}$
$\Rightarrow m=2 \Rightarrow P(0,1,2)$.
Let $\overrightarrow{P F}=\lambda \overrightarrow{P D}, \overrightarrow{A P}=(0,0,2)$,
$\overrightarrow{P D}=(\sqrt{3},-1,-2)$.
Then $\overrightarrow{A F}=\overrightarrow{A P}+\overrightarrow{P F}=(\sqrt{3} \lambda,-\lambda, 2-2 \lambda)$.
Let the normal vector of plane $P E C$ be $\boldsymbol{m}=(x, y, z)$,
$\overrightarrow{E P}=\left(\frac{\sqrt{3}}{2}, \frac{1}{2}, 2\right), \overrightarrow{P C}=(0,-2,-2)$.
Then $\left\{\begin{array}{l}\frac{\sqrt{3}}{2} x+\frac{1}{2} y+2 z=0, \\ -2 y-2 z=0\end{array}\right.$
$\Rightarrow \boldsymbol{m}=(-\sqrt{3},-1,1) \Rightarrow \boldsymbol{m} \cdot \overrightarrow{A F}=0$
$\Rightarrow-3 \lambda+\lambda+2-2 \lambda=0 \Rightarrow \lambda=\frac{1}{2}$
$\Rightarrow F$ is the midpoint of $P D$.
(2) By property (7), the equation of plane $P E A$ is
$$
x-\sqrt{3} y+\sqrt{3}=0 \text {, }
$$

the equation of plane $P E D$ is
$$
x+3 \sqrt{3} y+\sqrt{3} z=0 .
$$

Then the cosine value of the angle between the two normal vectors is
$$
\cos \theta=-\frac{4 \sqrt{31}}{31} \text {. }
$$

Therefore, the cosine value of the dihedral angle $D-P E-A$ is $\frac{4 \sqrt{31}}{31}$.