Soft SVM solving for $b$
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}&amp;\sum_{m \in S}  \alpha_m y^{(m)}\langle x^{(n)}, x^{(m)}\rangle\\&amp;=\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 \\
&amp;=\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\\
&amp;=\sum_{m=1}^N \alpha_m y^{(m)}\langle x^{(n)}, x^{(m)}\rangle \\
&amp;=\langle \sum_{m=1}^N \alpha_my^{(m)}x^{(m)}, x^{(n)}\rangle\\
&amp;= w^Tx^{(n)} \tag{3}\end{align}

Using $(2)$ and $(3)$, you should be able to obtain the conclusion.