Answer is (B).
Explanation: $T_1:r(P),r(Q),w(Q) T_2:r(Q),r(P),w(P).$
Now, consider any non serial schedule for example, $S:r_1(P),r_2(Q),r_1(Q),r2(P),w_1(Q),w_2(P).$
Now, draw a precedence graph for this schedule. Here there is a conflict from $T_1\to T_2$ and there is a conflict from $T_2\to T_1.$ Therefore, the graph will contain a cycle. So we can say that the schedule is not conflict serializable.