Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene and why? Is it (a) para-bromonitrobenzene and sodium methoxide or (b) sodium para-nitrophenoxide and bromomethane?

Or are they both appropriate? (And is one more appropriate than the other?)

One book (NCERT Chemistry Class 12 [India]) and at least one source Free NCERT solution of this question mention the latter as the appropriate. [However, I doubt the credibility of this source, as most likely, it is made to show how the NCERT book is correct (for CBSE Boards' exam) which, though is credible, is still a human-made book, which may have mistakes/errors.] NCERT only gives the answer as (b) without explanation. Another book (abc of Chemistry [India]) mentions "chemically, both are equally probable" as -nitro group activates the bromine site on benzene for substitution. And IMO, this book is more credible.

Any expert opinion?
In principle, both reactions will work.

(a) is a nucleophilic aromatic substitution, in which is $\ce{Br-}$ is displaced by $\ce{CH3O-}$

(b) is a methylation of a phenoxide

I'd probably go for (a) for two reasons:


The nitro group in para position decreases the charge density on the oxygen atom of the phenol. As a result, this anion will be less reactive than an unsubstituted phenoxide.
Methyl bromide with a boiling point of just 4 °C is difficult to handle and a poor choice for a methylation reagent. I'd rather try dimethyl sulfate if I had to.