I'm reading the book An Introduction to Operator Algebras (By Kehe Zhu). but do not know how to do the following exercise:
Let $A$ be the commutative C*-algebra $C(\partial D)$. For any $z\in \partial D$ let $\varphi_z$ be the point-evaluation on $z$. Show that $H_{\varphi_z}$ is one-dimensional. Show that there exists a countable subset $S'$ of $S(A)$ such that $$\Phi:A\rightarrow B\left(\bigoplus_{\varphi\in S'} H_\varphi\right)$$
as constructed in the proof of theorem 14.4 (Gelfand-Naimark-Segal Theorem) is a C*-isomorphism.
To do GNS, you take $C(\partial D)$ as your pre-Hilbert space, with the pre-inner product $$ \langle f,g\rangle=f(z)\overline{g(z)}. $$ Then you need to quotient by $$J_z=\{f:\ \varphi_z(f^*f)=0\} =\{f:\ |f(z)|^2=0\}=\{f: \ f(z)=0\}. $$ So we need to see that the quotient $C(\partial D)/J_z$ is one-dimensional. Now, $f-g\in J_z$ if and only if $g(z)=f(z)$, so the class of $f$ is determined by the value $f(z)$. That is, $$ C(\partial D)/J_z=\mathbb C+J_z, $$ and it is one-dimensional. Then its closure $H_{\varphi_z}$ will also be one-dimensional.
For the second part I will assume that $\partial D$ is the unit circle (if that's the case, $\mathbb T$ is a much more common notation). Let $S\subset\partial D$ be a dense subset. For each $s\in S$, consider the state $\varphi_s(f)=f(s)$. If we now consider the map $\Phi$ \begin{align} C(\partial D)&\longrightarrow B\left(\bigoplus_{s\in S}\mathbb C\right)=B\left(\bigoplus_{s\in S}H_{\varphi_s}\right)\\ f&\longmapsto \ \ \ \ \ M_f \end{align} where $M_f[g_s]_{s\in S}=[f(s)g_s]_{s\in S}$. This is clearly a $*$-homomorphism, and if $M_f=0$, then $f(s)=0$ for all $s\in S$ (by taking $g$ to be $1$ at $s$ and zero elsewhere), and so $f=0$ by continuity; thus $\Phi$ is a monomorphism. This map is not necessarily onto, but I think the book here is using "C$^*$-isomorphism" as a name for a $*$-monomorphism.
The set $S'$ is of course the set $S'=\{\varphi_s:\ s\in S\}$.