For R to be in BCNF, in all functional dependencies (X->Y), determinant must be a candidate key, i.e. X must be a candidate key (or, simply a key. Caution: It can not be a superkey, it must be a candidate key).

Now, in the question, there are three FDs (functional dependencies). Let's list them down: 1. ab -> cd 2. ac -> bd 3. ad -> b

In (1), ab is primary key (given) so it is a candidate key. So, (1) satisfies BCNF. In (2), ac can uniquely determine all of the rest of attributes, i.e. bd. So, it must be a candidate key as well. Therefore, (2) satisfies BCNF. In (3), ad is not a candidate key, because it does not uniquely determine rest of the attributes (it determines only b).

So, in (3) ad is not a candidate key, therefore BCNF is not satisfied.

Note, that R is in 3NF. 3NF holds if for every FD:X->A in the relation - (i) X is a superkey of R, or (ii) A is a prime attribute of R So, in (1) ab is the superkey, in (2) ac is the superkey, and in (3) b is a prime attribute (i.e. b is a subset of a candidate key). So, R is in 3NF.

The original question appears here - http://db.grussell.org/section009.html