The locus of the point of  intersection of  those normals to the parabola $x^2=8y$ which  are at right angles to each other is $x^2+ay+b=0$ then a+b 

 $x=my-4m-2m^3$
If  this passés  through p (h,k),we obtain 
$h=km-4m-2m^3$
This cubic has three roots ,say $m_1{m}_{2}and{m}_3$ Since two of the  normals are perpendicular we can  take $m_1{m}_2$ equal to -1 .From(1) we have 
Using $m_1{m}_2$ =-1  obtained ;
$h^2-2k+12=0$
Thus, the required  locus is   $x^2-2y+12=0$