If $A$ and $B$ are events such that $P(A\mid B)=P(B\mid A)$, then If $A$ and $B$ are events such that $P(A\mid B)=P(B\mid A)$, then

(A) $A \subset B$ but $A \neq B$

(B) $A=B$

(C) $A \cap B = \emptyset $

(D) $P(A)= P(B)$

My try

It is given that $P(A\mid B)=P(B\mid A)$

$ \implies \dfrac {P\left( A\cap B\right) }{P\left( B\right) } =\dfrac {P\left( B\cap A\right) }{P\left( A\right) }$

$\implies ( P\left( A\cap B\right))(P(A)-P(B))=0$

$\implies \  either \ P(A)=P(B) \ or \ P(A\cap B)=0$

$\implies \  either \ P(A)=P(B) \ or \ A\cap B=\emptyset$

But the correct answer is P(A)=P(B) 

Can anyone explain why $A\cap B=\emptyset$ is wrong?
Your logic implies, either $\mathbb{P}[A] = \mathbb{P}[B]$ or $\mathbb{P}[A \cap B] = 0$. Certainly, if $A \cap B = \emptyset$ the second condition holds, but consider $A \cap B$ being some discrete event out of a continuous probability, so $A \cap B \ne \emptyset$ but $\mathbb{P}[A\cap B] = 0$...

Thus it is possible (i.e. sufficient) that $A\cap B = \emptyset$ but it is not really necessary, and your question's style clearly implies a necessary condition...
Hello Girish Kumar chandora that's a good try but you just missed the point close
What you did is correct to an extent but you have taken 
 $P(A∩B)$as common but think about this
If it is given that $P(A∩B)/P(B)=P(B∩A)/P(A)$ 
Then the $P(A∩B)$ would be cancelled on both L.H.S and R.H.S as they both are same (basic rule in division)
So then u will be left behind with 
$P(A)=P(B)$