If $ P $ is equidistant from $ A $ and $ O $, it must lie on the perpendicular bisector of $ AO $. Since $ A $ has coordinates $ (10,-10) $ and $ O $ has coordinates $ (0,0) $, $ AO $ has slope $ \frac{-10-0}{10-0}=-1 $. The perpendicular bisector of $ AO $ must have slope $ -\frac{1}{-1}=1 $, and must also pass through the midpoint of $ AO $, which is $ (5,-5) $. Therefore, the perpendicular bisector has equation $ y-(-5)=x-5 $ or $ y=x-10 $. $ P $ is the point of intersection of the lines $ y=x-10 $ and the line $ y=-x+6 $. Setting these equations equal and solving for $ x $ yields $ -x+6=x-10 \Rightarrow x=8 $. It follows that $ y=-8+6=-2 $ and $ P=(x,y)=\boxed{(8,-2)} $.