From the resource allocation logic, it's clear that even numbered processes are taking even numbered resources and all even numbered processes share no more than $1$ resource. Now, if we make sure that all odd numbered processes take odd numbered resources without a cycle, then deadlock cannot occur. The "else" case of the resource allocation logic, is trying to do that. But, if $n$ is odd, $R_{n-i}$ and $R_{n-i-2}$ will be even and there is possibility of deadlock, when two processes requests the same $R_i$ and $R_j$. So, only $B$ and $D$ are the possible answers.
Now, in $D$, we can see that $P_0$ requests $R_0$ and $R_2$, $P_2$ requests $R_2$ and $R_4$, so on until, $P_{18}$ requests $R_{18}$ and $R_{20}$. At the same time $P_1$ requests $R_{40}$ and $R_{38}$, $P_3$ requests $R_{38}$ and $R_{36}$, so on until, $P_{17}$ requests $R_{24}$ and $R_{22}$. i.e.; there are no two processes requesting the same two resources and hence there can't be a cycle of dependencies which means, no deadlock is possible.
But for $B$, $P_8$ requests $R_8$ and $R_{10}$ and $P_{11}$ also requests $R_{10}$ and $R_8$. Hence, a deadlock is possible. (Suppose $P_8$ comes first and occupies $R_8$. Then $P_{11}$ comes and occupies $R_{10}$. Now, if $P_8$ requests $R_{10}$ and $P_{11}$ requests $R_8$, there will be deadlock)
Correct Answer: $B$
