Suppose $BP$ intersect at $AC$ at $M$. Then since $[ABP]=[BCP]$, we must have $M$ is the midpoint of $AC$. Thus $P$ must lie on $BM$ or the line through $B$ parallel to $AC$, and similarly for the other 3 sides. Let $N$ be the midpoint of $BD$. Now suppose $B,M,D$ lie on a straight line. Then $P$ must lie on $BD$, so $P$ must lie on $AN$ and $CN$, and the only possible point $P$ is $N$, and the condition for such a quad is $[ABD]=[BCD]$. Now suppose $B,M,D$ and $A,N,C$ are not collinear. Then $P$ cannot lie on both $BM,DM$. So WLOG $P$ lies on the line through $D$ parallel to $AC$ and on $BM$. Similarly WLOG $P$ lies on line through $C$ parallel to $BD$ and on $AN$. If $AC$ intersects $BD$ at $E$, then $DECP$ is a parallelogram. Then the condition for such a quad is $[ABCD]=2[DEC]$. The maximum number of such $P$ is 1, no matter the case.