A block A of mass m1 rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of the table and from its other end another block B of mass m2 is suspended. The coefficient of kinetic friction between the block and the table is μk. When the block A is sliding on the table, the tension in the string is:

The blocks m1 and m2 will move with combined acceleration a:
From F.B.D. of block m1
T−f1=m1a.. (i)
as the block m1 is sliding, kinetic friction will be acting:
T−μkN=m1a.. (ii)
N=mg.. (iii)
From F.B.D. of block m2
m2g−T=m2a.. (iv)
adding (ii) and (iv)
a=(m2g−μkN)/(m1+m2).. (v)
T=m2g−m2a=m2(g−(m2g−μkN)/(m1+m2))=m2((m1+m2)g−m2g+μkmg)/(m1+m2)=m2(m1g+μkm1g)/(m1+m2)=m1m2(1+μk)g/(m1+m2)
hence correct answer is option A