Since, $f$ is a polynomial time computable bijection and $f^{-1}$ is also polynomial time computable, $L_1$ and $L_2$ should have the same complexity (isomorphic). This is because, given a problem for $L_1$, we can always do a polynomial time reduction to $L_2$ and vice verse. Hence, the answer is 'C', as in 'A', $L_1$ and $L_2$ can be finite, in 'B', $L_1$ and $L_2$ can be in $P$ and in 'D', $L_1$ and $L_2$ can be recursive. Only, in 'C' there is no intersection for $L_1$ and $L_2$, and hence it canʼt be true.
Alternatively, we can prove 'C' to be false as follows:
Given $L_2$ is decidable. Now, for a problem in $L_1$, we can have a $TM$, which takes an input x, calculates $f(x)$ in polynomial time, check $f(x)$ is in $L_2$ (this is decidable as $L_2$ is decidable), and if it is, then output yes and otherwise no. Thus $L_1$ must also be decidable.