Assuming the applied force is exerted parallel to the floor, by Newton's second law both the net vertical and net horizontal forces would be zero:
∑ F (horizontal) = p - f = 0 … … … because the desk is pushed at a constant speed
∑ F (vertical) = n - mg = 0 … … … because the desk doesn't move up or down
where p is the magnitude of the applied force, f is the mag. of kinetic friction, n is the mag. of the normal force due to contact between the floor and desk, and mg is the weight of the desk. We have
n = mg = (100 kg) g = 980 N
and the mag. of friction is proportional to n according to
f = 0.4 n = 392 N
Then the applied force p has magnitude
p = f = 392 N