MISTAKE

Let $CE$ meet $AD$ at $F$.Now,$\frac{AF}{DF}=\frac{AC}{CD}\times\frac{\sin \angle ACF}{\sin \angle FCD}$.We have $\frac{AE}{BE}=\frac{AC}{BC}\times\frac{\sin \angle ACF}{\sin \angle FCD}$ or $\frac{\sin \angle ACF}{\sin \angle FCD}=\frac{BC}{AC}$.Therefore,$\frac{AF}{DF}=\frac{BC}{CD}=\frac{3}{1}$.

a solution by Stelios Karpathios from here

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Let $ AD \cap CE $, Using Barycentric Coordinates, It's easy to find, $$P\equiv (x:1:2) \text { and } P\equiv (1:1:z) \implies \boxed {P\equiv (1:1:2) \equiv (\frac {1}{4} , \frac {1}{4}, \frac {1}{2})} $$ WLOG, Let's assume $[ABC]=1$, We easily get, $$[APC]=\frac {1}{4} ,[ABD]=\frac {2}{3}, [APB]=\frac {1}{2}, [BPD]=\frac {1}{6},[BPC]=[BPD]+[PDC]=\frac {1}{6}+[PDC]=\frac {1}{4} \implies [PDC]=\frac {1}{12} $$ Hence, $$\frac{[APC]}{[PDC]}=\frac {AP}{DP}=\frac {\tfrac {1}{4}}{\tfrac {1}{12}}=3$$ Therefore, $$\boxed {AP:DP=3:1} $$