Proof. (1): By Theorem 34.9.2, $\norm{\phi}_{B^*}= \dpn{1_A, \phi}{B}$. By the Hahn-Banach Theorem, there exists $\Phi \in A^{*}$ such that $\Phi|_{B} = \phi$ and $\norm{\Phi}_{A^*}= \norm{\phi}_{B^*}= \dpn{1_A, \Phi}{A}$. Thus Theorem 34.9.2 implies that $\Phi \in S(A)$.

(2): Let $E(\phi) = \bracs{\Phi \in S(A)|\Phi|_B = \phi}$ be the collection of all extensions of $\phi$, then $E(\phi)$ is a weak*-closed convex subset of $S(A)$. By (1), $E(\phi)$ is non-empty, and as such admits an extreme point $\Phi$ by the Krein-Milman Theorem.

Let $\psi, \rho \in S(A)$ and $t \in (0, 1)$ such that $\Phi = (1 - t)\psi + t\rho$. In which case, $\phi = (1 - t)\psi|_{B} + t\rho|_{B}$. Since $\phi \in P(B)$, $\phi = \psi|_{B} = \rho|_{B}$, so $\psi, \rho \in E(\phi)$. As $\Phi$ is an extreme point of $E(\phi)$, $\Phi = \psi = \rho$. Therefore $\Phi \in P(A)$.$\square$