From $$y^{(n)}(w^T\cdot x^{(n)} + b)=1,$$
since $y^{(n)}$ is binary, we have $$w^T\cdot x^{(n)} + b = y^{(n)}$$
That is $$b=y^{(n)}-w^T\cdot x^{(n)}.\tag{2}$$
Now, let's examine
\begin{align}&\sum_{m \in S} \alpha_m y^{(m)}\langle x^{(n)}, x^{(m)}\rangle\\&=\sum_{m \in S} \alpha_m y^{(m)}\langle x^{(n)}, x^{(m)}\rangle + \sum_{m \notin S} 0\cdot y^{(m)}\langle x^{(n)}, x^{(m)}\rangle \\ &=\sum_{m \in S} \alpha_m y^{(m)}\langle x^{(n)}, x^{(m)}\rangle + \sum_{m \notin S} \alpha_m\cdot y^{(m)}\langle x^{(n)}, x^{(m)}\rangle\\ &=\sum_{m=1}^N \alpha_m y^{(m)}\langle x^{(n)}, x^{(m)}\rangle \\ &=\langle \sum_{m=1}^N \alpha_my^{(m)}x^{(m)}, x^{(n)}\rangle\\ &= w^Tx^{(n)} \tag{3}\end{align}
Using $(2)$ and $(3)$, you should be able to obtain the conclusion.