Answer is (B). Explanation: $T1:r(P),r(Q),w(Q) T2:r(Q),r(P),w(P)$ now, consider any non serial schedule for example, $S:r1(P),r2(Q),r1(Q),r2(P),w1(Q),w2(P)$ now, draw a precedence graph for this schedule. here there is a conflict from $T1->T2$ and there is a conflict from $T2->T1$ therefore, the graph will contain a cycle. so we can say that the schedule is not conflict serializable.