Let B be the subring of K containing A. I am going to prove that B is the localization of A at a prime ideal $P\subset A$, which seems a reasonable interpretation of the statement that B is a local ring of A.
First B is a local ring [by Atiyah-MacDonald Prop 5.18 i)] ; let $M_B$ be its maximal ideal. Similarly let $M_A$ be the maximal ideal of A. Define $P=A\cap M_B$. Then of course $P\subset M_A$ is prime and if we localize A at P, I claim that we get B:
a) If $\pi \in A-P$ then $\pi \notin M_B$ and so $\pi$ is invertible in B. Hence for all $ a \in A$ we have $a/ \pi \in B$. This shows $A_P \subset B$.
b) Now we see that B dominates $A_P$: this means we have an inclusion of local rings $A_P \subset B$ and of their maximal ideals: $PA_P \subset M_B$. But $A_P$ is a valuation ring of K because A is (This is the preceding exercise 28 : these guys know where they are heading!) and valuations rings are maximal for the relation of domination (Exercise 27: ditto !). Hence we have the claimed equality $B=A_P$.