Solution:

From the given, $A B=4 l$ and $B C=5 l$ for some constant $l>0$. Since $\angle A B C=60^{\circ}$, $B E=\frac{5 l}{2}$ and $C E=\frac{5 \sqrt{3} l}{2}$. Also, by the cosine law, $A C=\sqrt{21}$. Since $B E D C$ is cyclic, $\angle E D A=\angle A B C=60^{\circ}$. Consequently, $\angle E D B=30^{\circ}$ and $\triangle A E D \sim \triangle A C B$. From the latter, $A D=4 k$, $D E=5 k$, and $A E=\sqrt{21} k$ for some constant $k>0$. Since $4 l=A B=B E+A E=\frac{5 l}{2}+\sqrt{21} k$, then $\frac{l}{k}=\frac{2 \sqrt{21}}{3}$.

The area of $\triangle A B C$ equals
$$\frac{1}{2} \sin 60^{\circ} \cdot 4 l \cdot 5 l=\frac{1}{2} \cdot \sqrt{21} l \cdot 2 B M$$
which gives $B M=\frac{5 l}{\sqrt{7}}$. Observe that
$$\frac{C E}{D E}=\frac{5 \sqrt{3} l / 2}{5 k}=\frac{\sqrt{3} l}{2 k}=\frac{\sqrt{3}}{2} \cdot \frac{2 \sqrt{21}}{3}=\sqrt{7}=\frac{5 l}{5 l / \sqrt{7}}=\frac{C B}{M B}$$
This, along with $\angle M B C=\angle D B C=\angle D E C$, implies that $\triangle D E C \sim \triangle M B C$, so $\angle E C D=\angle B C M$ and thus, $\angle M C D=\angle B C E=30^{\circ}$. As $B M N C$ is cyclic, $\angle M B N=30^{\circ}$ so that lines $E D$ and $B N$ are parallel. We have $\angle D M C=60^{\circ}$ so that $\angle B P M=30^{\circ}$. Thus, $\triangle B M P$ is isosceles with $B M=M P$ and it follows that $M$ is the circumcenter of $\triangle B P D$. Therefore, $\angle B P D=90^{\circ}$. It follows that $\angle E D P=90^{\circ}$.