Some doubts:
Option B will be right option for it
Ck::ABD
PD=A->D,D->CE
So it is by default in 1NF
for converting it into 2NF we take closure of the A+ and D+
then we remove transitive dependency if any exist
then resulting table will be
R1(AC),R2(DCE) and R3(ABD) -->3NF
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