We have
$(D\to E), (\neg E \to A), (\neg B \to \neg C), (F \to \neg H), (\neg A \to B) $
Now, we have $(D,C)$ inside the hideout.
So, $E$ must also be inside (due to $D \to E),$ and $B$ must also be inside due to $([\neg B \to \neg C] \leftrightarrow [C \to B]).$ This eliminates options A, B and C.
Correct answer: D.