$B,D$ are equidistant from $AP$, so $AP$ must pass through the midpoint of $BD$. In the same way we conclude that $CP$ must pass through the midpoint of $BD$, while $BP,DP$ pass through the midpoint of $AC$. If $AP=CP$ or $BP=DP$ we're done: $AC$ or $BD$ is the diagonal we're looking for; if, on the other hand, we assume that $AP\ne CP$ and $BP\ne DP$, from the observation above it follows that $P$ is the common midpoint of the diagonals $AC,BD$, contradicting $AP\ne CP$.

My solution uses vectors. Here it is. Note that if we have shown that $P \in AC$ or $P \in BD$ then the conclusion follows. Then assume the contrary and note that the equality of the areas of $PAB$ and $PBC$ implies that $\vec{PA}\times\vec{PB}=\vec{PB}\times\vec{PC}$ or equivalently $(\vec{PA}+\vec{PC})\times \vec{PB}=0$. Thus $\vec{PA}+\vec{PC}=0$ which means that $P\in AC$ or $\vec{PA}+\vec{PC}$ and $\vec{PB}$ are collinear. In the same fashion we get that $\vec{PA}+\vec{PC}$ and $\vec{PD}$ are collinear and hence $\vec{PB}$ and $\vec{PD}$ are collinear which means $P\in BD$. In such a way we get a contradiction.