If $M$ is an $A$-module and $N$ is a $B$-module, then $Hom_A(B,M)$ is naturally a $B$-module (via the $B$-module structure on $B$ itself), and evaluation at $1 \in B$ gives an $A$-linear morphism $Hom_A(B,M) \to M$. This induces a morphism $Hom_B(N,Hom_A(B,M)) \to Hom_A(N,M)$, which you can easily check is an isomorphism.
Applying the preceding adjunction shows that $Hom_A(Hom_A(B,A),A) = Hom_B(Hom_A(B,A),Hom_A(B,A)) = End_B(Hom_A(B,A)),$ and so your question becomes the question of why the natural map $B \to End_B(Hom_A(B,A))$ is an isomorphism.
Now $B$ is finite over $A$, hence $Hom_A(B,A)$ is finite over $A$, thus also over $B$, and hence $End_B(Hom_A(B,A))$ is a finite $B$-algebra. We will show that it equals $B$ by showing that it embeds into the fraction field of $B$ (and then appealing to the integral closedness of $B$).
If we write $K$ and $L$ for the fraction fields of $A$ and $B$ respectively, then $K\otimes_A B = L$, and so $Hom_A(B,A) \hookrightarrow Hom_A(B,K) = Hom_K(L,K)$. The last of these spaces is one-dimensional over $L$, and hence $End_L(Hom_K(L,K)) = L.$
From the results of the preceding paragraph, we can draw some conclusions:
The natural map $Hom_A(B,A) \to L\otimes_B Hom_A(B,A)$ is injective, i.e. $Hom_A(B,A)$ is torsion-free over $B$, and hence $End_B(Hom_A(B,A))$ is torsion-free over $B$ as well, i.e. the natural map $End_B(Hom_A(B,A)) \to L\otimes_B End_B(Hom_A(B,A))$ is also injective.
$L\otimes_K Hom_A(B,A) = Hom_K(L,K)$, and hence $L\otimes_B End_B(Hom_A(B,A)) = End_L(Hom_K(L,K)) = L.$
From 1. and 2. we see that $End_B(Hom_A(B,A))$ is a subalgebra of $L$ containing $B$. It is finite over $B$ (as noted above) and so equals $B$.
This completes the argument. Note that I didn't use the assumption that $A$ is integrally closed, so either the statement is true without this hypothesis, or I made a mistake. I wouldn't rule out the latter possibility!