Correct Option: A

First (S) = First (C) = {c,d} there are no multiple in single row of parsing table hence grammar is LL₁
Note : if we have A → B/C for grammar to be LL(1) first(B) intersection First(C) should be null otherwise grammar is not LL₁. If First(B) contains epsilon then Follow(A) intersection First(C) should be null. Using this we can say grammar is LL(1) or not without constructing parsing table.
An ∈ ∈ free LL(1) grammar is also SLR(1) and hence LALR(1) and LR(1) too.
Since there is no conflict, the grammar is LL(1). We can construct a predictive parse table with no conflicts. This grammar also LR(0), SLR(1), CLR(1) and LALR(1).