2020 AMC 10A Problem 18

Only parity matters. Modulo $2$, the condition is that $ad-bc$ is $1$, meaning the matrix $\begin{pmatrix}a&b\\ c&d\end{pmatrix}$ is invertible over $\mathbb F_2$. There are $(4-1)(4-2)=6$ invertible $2\times2$ matrices over $\mathbb F_2$. Each parity pattern lifts to $2^4=16$ choices from $\{0,1,2,3\}$, so there are $6\cdot16=96$ quadruples. Thus, C is the correct answer.