Czech-Polish-Slovak Match

If $P$ lies on one of the diagonals $AC$ or $BD$, let's say on $AC$, then $$\frac{[PAB]}{[PBC]}=\frac{AP}{PC}=\frac{[PDA]}{[PCD]},$$ which is the desired equality. We prove that no other point lying inside $ABCD$ satisfies the conditions of the problem.

Denote by $O$ the point of the intersection of the diagonals $AC$ and $BD$ and suppose that $P$ lies inside the triangle $ABO$. Let moreover $BP$ and $AC$ meet at $Q$ and $DP$ and $AC$ meet at $R$. Then $$\frac{[PAB]}{[PBC]}=\frac{AQ}{QC}\text{and}\frac{[PDA]}{[PCD]}=\frac{AR}{RC},$$ which, since $Q\ne R$, implies that the given equality cannot hold.

Therefore the desired locus of the points $P$ consists of the diagonals $AC$ and $BD$.