19. (12 points) As shown in Figure 2, for the pyramid $P-ABCD$:
(1) Find a point $F$ on edge $PD$ such that $AF \parallel$ plane $PEC$;
(2) Find the cosine value of the dihedral angle $D-PE-A$. 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 between $PC$ and plane $PAD$ is $\frac{\sqrt{6}}{4}$.
19. (1) Taking $BD$ as the $x$-axis, $CA$ as the $y$-axis, and the intersection of $AC$ and $BD$ as $O$, and drawing a perpendicular line to the plane $ABCD$ through point $O$ as the $z$-axis, we establish a spatial rectangular coordinate system.
Thus, $A(0,1,0), B(-\sqrt{3}, 0,0), C(0,-1,0)$, $D(\sqrt{3}, 0,0), P(0,1, m), E\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}, 0\right)$.
Therefore, $\overrightarrow{P C}=(0,-2,-m)$.
Let the normal vector of plane $PAD$ be $\boldsymbol{n}=(x, y, z)$,
$\overrightarrow{A P}=(0,0, m), \overrightarrow{A D}=(\sqrt{3},-1,0)$.
Thus, $\left\{\begin{array}{l}m z=0, \\ \sqrt{3} x-y=0\end{array} \Rightarrow \boldsymbol{n}=(\sqrt{3}, 3,0)\right.$
$\Rightarrow|\cos \langle\overrightarrow{P C}, \boldsymbol{n}\rangle|=\left|\frac{-6}{\sqrt{4+m^{2}} \sqrt{12}}\right|=\frac{\sqrt{6}}{4}$
$\Rightarrow m=2$.
Therefore, $P(0,1,2)$.
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 $PEC$ be $\boldsymbol{m}=(x, y, z)$,
$\overrightarrow{E P}=\left(\frac{\sqrt{3}}{2}, \frac{1}{2}, 2\right), \overrightarrow{P C}=(0,-2,-2)$.
By $\left\{\begin{array}{l}\frac{\sqrt{3}}{2} x+\frac{1}{2} y+2 z=0 \\ -2 y-2 z=0\end{array}\right.$,
$\Rightarrow \boldsymbol{m}=(-\sqrt{3},-1,1)$.
Also, $\boldsymbol{m} \cdot \overrightarrow{A F}=0$
$\Rightarrow-3 \lambda+\lambda+2-2 \lambda=0$
$\Rightarrow \lambda=\frac{1}{2}$.
Therefore, $F$ is the midpoint of $PD$.
(2) Let the normal vectors of plane $PEA$ and plane $PED$ be $\boldsymbol{n}_{1}=(\sqrt{3},-3,0), \boldsymbol{n}_{2}=(\sqrt{3}, 9,-3)$.
Then $\cos \left\langle\boldsymbol{n}_{1}, \boldsymbol{n}_{2}\right\rangle=\frac{3-27}{\sqrt{12} \sqrt{93}}=-\frac{4 \sqrt{31}}{31}$. Therefore, the required cosine value is $\frac{4 \sqrt{31}}{31}$.