R( A, B, C, D, E, F).
Given FDs are - AB - > CD, E -> C, B -> EF.
Now check option A.
We are decomposing R into 3 relation R1(A, B, C, D), R2(C,E), and R3(E, F, B).
where DFs for R1 : AB -> CD
and DFs for R2 : E -> C
and DFs for R3 : B -> EF.
Here all FDs are preserved and all decomposition are lossless too( because common attribute between every decomposed table is a C.K. in one of the table. E.g. - common attribute between R2 and R3 is E which is a key in R2 table...similarly you can check for others). And FDs of all three decomposed relation contains super key in the left side of FDs. So all decomposed relation are in 3NF and even in BCNF.
Now check option B.
Here we are decomposing R into 3 relation R1(A, B, D), R2(E,C), and R3(E, F, B).
where DFs for R1 : AB -> D
and DFs for R2 : E -> C
and DFs for R3 : B -> EF.
Here it seems like AB -> CD is not preserved but we can get AB -> C also with the help of other FDs as B -> EF and now from E -> C. So here also all the FDs are preserved and FDs of all three decomposed relation contains super key in the left side of FDs. So all decomposed relation are in 3NF and even in BCNF.
Now check option C.
Here we are decomposing R into 3 relation R1(A, B,C, D), R2(C,E), and R3( B,E,F).
Same as option A.
Now check option D.
Here we are decomposing R into 3 relation R1(A, B,C, D,E,F), R2(E,C), and R3(E, F, B).
where DFs for R1 : AB -> CD, E->C, B->EF
and DFs for R2 : E -> C
and DFs for R3 : B -> EF.
Now if we see relation R1 where C.K is B and its FDs are-
AB -> CD ( left side is a super key ..so no probem)
E -> C ( neither left side contains super key nor right side contains prime attribute so it violates the 3NF rule )
B -> EF ( left side is a super key ..so no probem ).
Ans - A, B, C