Let $BD$ and $CD$ meet $AC$ and $AD$ at $X$ and $Y$ resp.
lets define $\angle ACE = \beta$, $\angle DCE = \alpha$, $\angle BDA = \theta$, $\angle CDB = \gamma$
Now, we can easily get that $\bigtriangleup ABD \cong \bigtriangleup ACE$ and $\bigtriangleup BCD \cong \bigtriangleup CDE$
this gives us that $ABCP$ and $APDE$ is concyclic. So, $\angle CBD = \angle CAM = \alpha$ and $\angle DAM = \gamma$. Now.sine law on $\bigtriangleup CDY$ and $\bigtriangleup CYA$ gives us
$\dfrac{DY}{YA} = \dfrac{\sin \alpha \sin (\alpha + \gamma)}{\sin \beta \sin(\gamma + \theta)}$
Again, sine law on $\bigtriangleup CXB$ and $\bigtriangleup ABX$ gives us
$\dfrac{AX}{XC} = \dfrac{\sin \beta \sin(\gamma + \theta)}{\sin \alpha \sin (\alpha + \gamma)}$
Now, let $AP \cap CD = M$. Ceva's theorem on $\bigtriangleup ACD$ gives us $\dfrac{CM}{MD} = 1$, or,
$CM = MD$