Let $\mathcal{A}$ be a unital separable $C^*$-algebra. Let $\mathcal{A}^{**}$ be the double dual space of $\mathcal{A}$. Let $p$ be a central support projection in $\mathcal{Z}(\mathcal{A}^{**})$. I want to claim that the state space of $p\mathcal{A}^{**}$, $\mathcal{S}\left(p\mathcal{A}^{**}\right)$ is a metrizable space.
Using Goldstine Theorem, we obtain that under the canonical embedding $j:\mathcal{A}\to\mathcal{A}^{**}$, $j(A)$ is a weak$^*$-dense subset of $\mathcal{A}^{**}$. Since $\mathcal{A}$ is separable (say $ \{a_i: i\in \mathbb{N}\}$ is a countable dense subset), we obtain that $\{j(a_i): i \in \mathbb{N}\}$ is a countable weak$^*$-dense subset of $\mathcal{A}^{**}$. In particular, we see that $\{pj(a_i): i \in \mathbb{N}\}$ is a countable weak$^*$-dense subset of $p\mathcal{A}^{**}$. We can now effectively define metric $d$, $$d(\varphi,\phi)=\sum_{n=1}^{\infty}\frac{1}{2^n}|\varphi(pj(a_i)-\phi(pj(a_i)|, ~\varphi,\phi\in \mathcal{S}\left(p\mathcal{A}^{**}\right) .$$
Is this correct? Is there any other way to see the claim? Thank you for the help.
Yes, I think the argument is fine. You need to consider your countable dense subset in the unit ball, though; but it doesn't hurt your argument. The result is not surprising because the price you are paying is going from the norm topology to the weak$^*$-topology.