19. 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 between $PC$ 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 $D-PE-A$.
(1) As shown in the figure, $G, F$ are the midpoints of $P C, P D$ respectively, then $G F \cong \frac{1}{2} C D \cong E A \Rightarrow$ quadrilateral $A F G E$ is a parallelogram, so $A F / / E G \Rightarrow A F / /$ plane $P E C$.
(2) Let $M$ be the midpoint of $A D$, then $\triangle A C D$ is an equilateral

triangle $\Rightarrow C M=\sqrt{3}$ and $C M \perp$ plane $P A D$,
thus $\angle M P C$ is the angle between $P C$ and plane $P A D$
$$
\begin{array}{l}
\Rightarrow \tan \angle M P C=\frac{C M}{P M}=\frac{\sqrt{3}}{\sqrt{5}} \\
\Rightarrow P M=\sqrt{5}, P A=2 .
\end{array}
$$

Draw $D H \perp A B$, draw $H N \perp P E$, and connect $D N$. Then $D H \perp$ plane $P A B \Rightarrow \angle H N D$ is the plane angle of the dihedral angle $D-P E-A$, $\tan \angle P E A=2 \Rightarrow H N=2 \sin \angle P E A=\frac{4}{\sqrt{5}}, H D=\sqrt{3}$ $\Rightarrow \tan \angle H N D=\frac{\sqrt{15}}{4} \Rightarrow \cos \angle H N D=\frac{4}{\sqrt{31}}=\frac{4 \sqrt{31}}{31}$.
Therefore, the cosine value of the dihedral angle $D-P E-A$ is $\frac{4 \sqrt{31}}{31}$.