(First, some notation. Each of the following summations is over all k. I'll use the & symbol to denote an intersection of sets.)
It sounds like you're happy with the "CT3 version" of the Law of Total Probability. This states that, for any event X (and sets A_k as you've already defined),
P(X) = ΣP(X|A_k)P(A_k)
... = ΣP(X&A_k).
Putting X = B&C gives
P(B&C) = ΣP(B&C&A_k). <-- (1)
By definition,
P(B|C) = P(B&C) ÷ P(C).
Using (1),
P(B|C) = [ ΣP(B&C&A_k) ] ÷ P(C)
... = Σ[ P(B&C&A_k) ÷ P(C) ]
... = Σ[ [P(B&C&A_k)÷P(C&A_k)] * [P(C&A_k)÷P(C)] ]
... = Σ[ P(B|C&A_k) * P(A_k|C) ],
which is the result we were looking for. I'll admit that I've missed out some of the technicalities (e.g. P(C) non-zero) but otherwise I think this works.
I would've thought the proof of this result was unlikely to come up. The author of the Core Reading seems to regard it as an "obvious" (hmm) extension of the CT3 formula, with a "|C" tacked on the end of everything. So it may be safe to just state it.
But I guess you never know what the Examiners want to throw at you! I hope this helps.