Consider R(ABCD) {AB-->CD, C-->D, B-->D} and AB is the primary key. Every non-key (C and D) is functionally determined by primary key (AB) [condition satisified].
You will argue that C-->D (transitive) and B-->D (partial) are invalid and violate the condition. But question did not say that D must be determined "ONLY" by a primary key. Once we satisfy the condition, we are free to see more possibilities try to prove 2NF and 3NF false.
And if you see trivially, primary key ALWAYS determine EVERY non-key attribute by default. So the statement really does not add any useful information about the possible FD set that can satisfy higher normal forms. It may or may not satisfy. Our FD set in example SATISFIES the condition and it is just in 1NF.
But yes, if every non-key attribute is determined "ONLY" by the primary key, then it will be in minimum 3NF as there will not be any partial / transitive dependency then.
Hence the answer in this case will be 1NF. But as the answer key suggests 3NF, they really must be considering the latter case and framed the question incorrectly.