Suppose that $\phi$ is a positive linear functional on $C^{*}(\Gamma)$. Let $\Gamma$ be a discrete abelian group. We denote the group ring of $\Gamma$ by $\mathbb{C}(\Gamma)$, which is the set of all formal sums of the form $\sum_{s\in \Gamma}a_s s$, where only finitely many of the scalar coefficients $a_s \in \mathbb{C}$ are nonzero. The full(or universal) group $C^*$-algebra of $\Gamma$, denoted by $C^{*}(\Gamma)$ is the completion of $\mathbb{C}(\Gamma)$ with respect to the norm $$||x||_u=\sup_{u}||\pi(x)||$$ Now suppose that $\phi$ is a linear positive functional on $\Gamma$: For $s_1,s_2,\ldots,s_n \in \Gamma$ we have $$[\phi(s_i^{-1}s_j)]_{i,j}=(\text{id}_n \otimes\phi)\left(\begin{bmatrix}s_1 &s_2 &\ldots&s_n\\0 & 0& \ldots &0\\\vdots&\vdots &\ddots &\vdots \\ 0 & 0 & \ldots &0\end{bmatrix}^{*}\begin{bmatrix}s_1 &s_2 &\ldots&s_n\\0 & 0& \ldots &0\\\vdots&\vdots &\ddots &\vdots \\ 0 & 0 & \ldots &0\end{bmatrix} \right)$$ which is positive since $\phi$ is a completely positive map.
Let $C_c(\Gamma)$ be the finitely supported functions on $\Gamma$. Define a sesquilinear form $C_c(\Gamma) \times C_c(\Gamma) \to \mathbb{C}$ by $$\langle f,g\rangle_{\phi}=\sum_{s,t \in \gamma}\phi(s^{-1}t)f(t)\bar{g(s)}$$ This form is positive semidefinite: For $f \in C_c(\Gamma)$ has support $F$, then $$\langle f, f \rangle_{\phi}=\sum_{s,t \in \Gamma}\phi(s^{-1}t)f(t)\bar{f(s)}=\langle[\phi(s^{-1}t)]_{s,t \in F}(f) ,f \rangle =\langle[\phi(s^{-1}t)]^{\frac{1}{2}}_{s,t \in F}(f) ,[\phi(s^{-1}t)]^{\frac{1}{2}}f \rangle \ge 0$$
Let $N=\{f \in C_c(\Gamma):\langle f,f\rangle_{\phi}=0\}$. Then Let $\cal{l}^2_{\phi}(\Gamma)$ be the Hilbert completion of $C_c(\Gamma)/N$. This is all the process of creation of the space.
Show that the GNS space of $C^*(\Gamma)$ with respect to $\phi$ is nothing but $l^2_{\phi}(\Gamma)$.
Let $M=\{f \in C^{*}(\Gamma):\phi(f^*f)=0\}$. Then the GNS construction of $C^*(\Gamma)$ (let's denote by $\bar{C}^*(\Gamma)$) is the Hilbert completion of $C^*(\Gamma)/M$. I want to define an isometric isomorphism from $\bar{C}^*(\Gamma)$ to $l^2_{\phi}(\Gamma)$. It is enough to define the map between dense subsets. If I look at a typical element $\sum_{s \in \Gamma}a_s s$ in $\mathbb{C}(\Gamma)$, then I can define $f \in C_c(\Gamma)$ corresponding to it by $ f(t) = a_t $ (This is because $a_t \ne 0$ for finitely many $t \in \Gamma$). Let $$T: \mathbb{C}(\Gamma)/M \to C_c(\Gamma)/N$$ by $$T\left(\sum_s a_s s +M\right)= f+N$$, where $f$ is defined as above.
I don't know if this will work or not. Thanks for the help!!
Yes, your construction works.
To see this, first observe that $T$ is an isometry:
If $f=\sum_{g\in G} a_gg$, finitely supported, we get $$||f+N||_{\varphi}^2=\langle f+N,f+N\rangle_{GNS}=\varphi(f^*f)=\sum_{g,s\in \Gamma}a_g\overline{a_s}\varphi(g^{-1}s)=\langle \hat{f},\hat{f}\rangle_{\varphi}=||\hat{f}||^2$$
This implies that $T$ is well-defined and isometric.
So, we can extend it to an isometric map, also denoted by $T$, from $l_{\varphi}^2(\Gamma)$ to $H_{\varphi}$.
It is easy to check that $T$ is surjective.