[Solution] A point and a parallelogram

 The problem was here: https://ruffnsluff.blogspot.com/2021/08/a-point-and-parallelogram.html

We use the following facts

1) The diagonals of a parallelogram bisect each other.

2) Apollonius theorem: If $A,B,C$ is a triangle and $D$ is midpoint of $BC$, then $|AB|^2 + |AC|^2 = 2(|AD|^2 + |BD|^2)$

 Let $O$ be the intersection of the diagonals $AC$ and $BD$ of the parallelogram $ABCD$.

The by applying the Apollonius theorem of triangle $PAC$ we get that 

$$|PA|^2 + |PC|^2 = 2(|AO|^2 + |PO|^2)$$

Similarly

$$|PB|^2 + |PD|^2 = 2(|BO|^2 + |PO|^2)$$

Adding we get

$$|PA|^2 + |PB|^2 + |PC|^2 + |PD|^2 = 4|PO|^2 + 2(|AO|^2 + |BO|^2)$$

Thus if the LHS is constant then so is |PO| and hence the locus of $P$ is a circle with center $O$ (or empty if the constant is smaller than $2(|AO|^2 + |BO|^2)$).