14. In the quadrilateral pyramid $P-ABCD$, the side face $PAD$ is an equilateral triangle, and it is perpendicular to the base $ABCD$, the base $ABCD$ is a rhombus with a side length of 2, $\angle BAD=60^{\circ}$, $N$ is the midpoint of $PB$, and the section $DAN$ intersects $PC$ at $M$. Find the size of the dihedral angle $P-AD-M$.
14. Solution 1 As shown in Figure (a), take the midpoint $O$ of $A D$, and connect $P O, B O, N O$.

Since $\triangle A B D$ is an equilateral triangle, $A D \perp B O$. Similarly, $A D \perp P O$.
Therefore, $A D \perp$ plane $P B O$, so $A D \perp P B$.
Also, $P A=A B=2$, and $N$ is the midpoint of $P B$, so $A N \perp P B$, which means $P B \perp$ plane $A D M N$.
Connect $O N$, then $O N$ is the projection of $P O$ on plane $A D M N$,
By the inverse of the three perpendiculars theorem, $A D \perp N O$, hence $\angle P O N$ is the plane angle of the dihedral angle $P-A D-M$.
The lateral face $P A D \perp$ the base face $A B C D$, and $P O \perp A D$, so $P O \perp$ plane $A B C D$.
It is easy to see that $P O=B O$, in the isosceles right $\triangle P O B$, $N$ is the midpoint of the hypotenuse, so $\angle P O N=45^{\circ}$, which means the dihedral angle $P-A D-M$ is $45^{\circ}$.
Solution 2 As shown in Figure (b), take the midpoint $O$ of $A D$, and connect $P O, B O, N O$.

Since $\triangle A B D$ is an equilateral triangle, $A D \perp B O$.
Similarly, $A D \perp P O$.
The lateral face $P A D \perp$ the base face $A B C D$, so $P O \perp$ plane $A B C D$.
Establish a spatial rectangular coordinate system with $O$ as the origin as shown in the figure,
Then $O(0,0,0), B(0, \sqrt{3}, 0), P(0,0, \sqrt{3}), A(1,0,0), D(-1,0,0), N\left(0, \frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}\right)$.
Therefore, $\overrightarrow{A D}=(-2,0,0), \overrightarrow{B P}=(0,-\sqrt{3}, \sqrt{3}), \overrightarrow{A N}=\left(-1, \frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}\right)$.
Therefore, $\overrightarrow{A D} \cdot \overrightarrow{B P}=0, \overrightarrow{A N} \cdot \overrightarrow{B P}=0$, which means $\overrightarrow{B P}$ is a normal vector of plane $A D M N$.
It is easy to see that $\overrightarrow{B O}$ is a normal vector of plane $P A D$.
Thus, $\cos \langle\overrightarrow{B P}, \overrightarrow{B O}\rangle=\frac{\overrightarrow{B P} \cdot \overrightarrow{B O}}{|\overrightarrow{B P}| \cdot|\overrightarrow{B O}|}=\frac{3}{\sqrt{3} \cdot \sqrt{6}}=\frac{\sqrt{2}}{2}$.
Therefore, the dihedral angle $P-A D-M$ is $45^{\circ}$.