The given boolean gate $D$ is the implication gate.
Implication gate is functionally complete (with the help of boolean input $0$). Proof HERE.
So, {$D,0$} is functionally complete, hence, we can realize EVERY switching function using these.
So, the answer is $A,C.$
Implementing NOT gate $\bar{B}$ using {$D,0$}: Make input $A = 0.$
Detailed Video Explanation: https://www.youtube.com/watch?v=RH-H5fRiewQ