Let distance between A and B be d
Then potential at any distance x from point A is given by
V=kqx+kq(d−x)
V=kq(1x+1(d−x))
Now
dV/dx=kqd(2x−d)(x(d−x))2
therefore for x<d/2
dV/dx<0 therefore potential decreases
and for x>d/2
dV/dx>0 therefore potential increases
Hence first decreases then increases option (D)