Consider the first FD $:N\to W$
$N$ is not a superkey of the complete relation, so it violates BCNF.
Decompose into
$R_1(N,W)$ and $R'(I,N,S,Q,T,P,R)$.
$R_1$ is in BCNF because $N$ is its key.
Now examine $R'$.
The FD $IN\to S$ still holds.
But $IN$ does not determine $Q,T,P,R$, so $IN$ is not a superkey of $R'$.
Thus it also violates BCNF.
Decompose $R'$ into
$R_2(I,N,S)$ and $R_3(I,N,Q,T,P,R)$.
In $R_2$, $IN\to S$, so $IN$ is a key.
$R_3$ has no non-trivial projected FD from the given set.
Therefore, the final BCNF decomposition is $\boxed{{NW,\ INS,\ INQTPR}}$
Answer : A
