The vector BC+BA is equal to the diagonal of the parallelogram with BC and BA passing through point B (assuming BD), and AC is the other diagonal of the parallelogram.

So BP = BD/2. That is, BP is half of diagonal BD, so point P must be the intersection of two diagonal lines and the bisector of the two diagonal lines.

In other words, point P is also the midpoint of AC.

Then, considering the directivity PA = -PC, there must be a vector PC+PA = 0.

So, the correct answer is B.