10. (15 points) As shown in Figure 1, in the pyramid $P-ABCD$, the base $ABCD$ is a rhombus, $\frac{AB}{AC}=\frac{2}{3}$, $PA \perp$ plane $ABCD$, and $M$ is a point on edge $PC$.
(I) If the angles formed by $PA$, $PB$, $PC$, $PD$ with plane $ABCD$ are $\alpha$, $\beta$, $\gamma$, $\delta$ respectively, and $\alpha=\beta+\gamma+\delta$, find $\sin (\beta+\gamma)$;
(2) If $PA=2AC$, when point $M$ makes the area of $\triangle MBD$ the smallest, find the size of the dihedral angle $M-BD-A$.
10. (1) As shown in Figure 5, since $P A \perp$ plane $A B C D$, we know $\alpha=90^{\circ}$, and $A B, A C, A D$ are the projections of $P B, P C, P D$ in plane $A B C D$, respectively. Therefore,
$$
\angle P B A=\beta, \angle P C A=\gamma, \angle P D A=\delta.
$$

Since quadrilateral $A B C D$ is a rhombus, we have $A B=A D$.
Also, $P A=P A$, so
Rt $\triangle P A B \cong$ Rt $\triangle P A D$.
Thus, $\beta=\delta$.
Since $\beta+\gamma+\delta=\alpha=90^{\circ}$, we have $2 \beta+\gamma=90^{\circ}$.
Therefore, $\tan 2 \beta=\cot \gamma$.
Let $P A=x, A B=2 a, A C=3 a$. Then
$\tan \beta=\frac{P A}{A B}=\frac{x}{2 a}, \cot \gamma=\frac{A C}{P A}=\frac{3 a}{x}$.
Thus, $\frac{2 \cdot \frac{x}{2 a}}{1-\left(\frac{x}{2 a}\right)^{2}}=\frac{3 a}{x}$.
Solving this, we get $\tan \beta=\frac{x}{2 a}=\frac{\sqrt{21}}{7}$.
Since $2 \beta+\gamma=90^{\circ}$, we have
$\sin (\beta+\gamma)=\cos \beta=\frac{A B}{P B}=\frac{2 a}{\sqrt{(2 a)^{2}+x^{2}}}$
$=\frac{1}{\sqrt{1+\left(\frac{x}{2 a}\right)^{2}}}=\frac{\sqrt{70}}{10}$.
(2) Let $A C \cap B D=O$, and connect $O M$. By symmetry, we have $M B=M D$. Therefore, $O M \perp B D$, meaning $O M$ is the altitude of isosceles $\triangle M B D$, and the length of the base $B D$ is constant. Thus, the area of $\triangle M B D$ is minimized when $M O$ is minimized, which is equivalent to $O M \perp P C$. At this point,
$$
\angle M O C=90^{\circ}-\angle O C M=\angle A P C.
$$

Since quadrilateral $A B C D$ is a rhombus, we have $A O \perp B D$.
Also, $O M \perp B D$, so $\angle M O A$ is the plane angle of the dihedral angle $M B D-A$, and we have
$$
\begin{array}{l}
\cos \angle M O A=-\cos \angle M O C=-\cos \angle A P C \\
=-\frac{P A}{P C}=-\frac{P A}{\sqrt{P A^{2}+A C^{2}}} \\
=-\frac{1}{\sqrt{1+\left(\frac{A C}{P A}\right)^{2}}}=-\frac{2 \sqrt{5}}{5}.
\end{array}
$$

Therefore, $\angle M O A=\pi-\arccos \frac{2 \sqrt{5}}{5}$.