Let $ABCD$ be a cyclic quadrilateral. Let $M, N$ be the midpoints of the diagonals $AC$ and $BD$ and let $P$ be the midpoint of $MN$. Let $A',B',C',D'$ be the intersections of the rays $AP$, $BP$, $CP$ and $DP$ respectively with the circumcircle of the quadrilateral $ABCD$. Find, with proof, the value of the sum \[ \sigma = \frac{ AP}{PA'} + \frac{BP}{PB'} + \frac{CP}{PC'} + \frac{DP}{PD'} . \]