k = QA^2+QB^2+QC^2 = 32, where Q is the barycentre (3, 1) of ABC.
To see this, replace PA by PG+QA etc., expand the squares and use distributivity and the defining property of the barycenter QA+QB+QC=0.
You can also let P be (x, y) and check that PA^2+PB^2+PC^2 = (x-4)^2+(y+1)^2 + … can be transformed into 3[(x-3)^2 + (y-1)^2] + 32.