In parallelogram $ABCD$, point $P$ is on diagonal $\overline{AC}$ such that $\overline{AP} : \overline{PC} = 2 : 1$. The area of $\triangle ABD$ is $54$ square units. Find the area of $\triangle APD$. Express your answer in simplest radical form. Let's think step by step and output the final answer within \boxed{}.
user