B′ will have the elements which are outside the circle drawn for B
⇒B′= { a,b,c,i,k,l,m,n,p }
(A∩B′) will have the elements which are common among the elements written inside the circle of A and in B′= { a,b,c,i,k,l,m,n,p }.
⇒(A∩B′)= { a,b,c,i } .....(1)
And A−B will have the elements inside the circle drawn for A but not inside the circle drawn for B.
So, A−B= { a,b,c,i } .....(2)
From (1) and (2), we have
A∩B′=A−B