Separating pure states on the $2\times 2$ matrix algebra

The answer to my question is no. Let $\mathcal{A} = M_2$ be the algebra of $2\times 2$ complex matrices, let $\mathcal{B}$ be the subalgebra of diagonal matrices, let $\mathcal{A}'$ be a C*-algebra which unitally contains $\mathcal{A}$, and let $\mathcal{B}'$ be a C*-subalgebra of $\mathcal{A}'$ which contains $\mathcal{B}$.

$\mathcal{A}$ contains the matrix units $e_{11}$, $e_{12}$, $e_{21}$, $e_{22}$, and hence so does $\mathcal{A}'$. So we have $\mathcal{A}' \cong M_2\otimes \mathcal{C}$ for some other unital C*-algebra $\mathcal{C}$. Also, since $\mathcal{B}$ contains $e_{11}$ and $e_{22}$, we have $$\mathcal{B}' = e_{11}\mathcal{B}'e_{11} + e_{11}\mathcal{B}'e_{22} + e_{22}\mathcal{B}'e_{11} + e_{22}\mathcal{B}'e_{22} = \mathcal{B}'_{11} + \mathcal{B}'_{12} + \mathcal{B}'_{21} + \mathcal{B}'_{22}.$$ We have $P = e_{11} + e_{12} + e_{21} + e_{22}$. There are two cases.

Case 1. $\mathcal{B}'_{12}$ contains an element $x$ satisfying $\|x - e_{12}\| < 1$. Then $\mathcal{B}'$ contains the element $y = e_{11} + x + x^* + e_{22}$, and we have $$\|y - P\| = \|(x - e_{12}) + (x^* - e_{21})\| = \|x - e_{12}\| < 1.$$ So in this case the distance from $\mathcal{B}'$ to $P$ is strictly less than 1.

Case 2. We have $\|x - e_{12}\| \geq 1$ for all $x \in \mathcal{B}'_{12}$. By the Hahn-Banach theorem, we can find a norm one linear functional on $e_{11}\mathcal{A}'e_{22}$ which takes $e_{12}$ to $1$ and vanishes on $\mathcal{B}'_{12}$. By the isomorphism $e_{11}\mathcal{A}'e_{22} \cong \mathcal{C}$, this yields a state $h$ on $\mathcal{C}$. Then $f\otimes h$ and $g\otimes h$ are two states on $\mathcal{A}'$ which respectively extend $f$ and $g$, but $\mathcal{B}'$ does not separate them, because they agree on $\mathcal{B}'_{11} \subseteq e_{11}\mathcal{A}'e_{11}$ and $\mathcal{B}'_{22} \subseteq e_{22}\mathcal{A}'e_{22}$, and they both vanish on $\mathcal{B}'_{12}$ and $\mathcal{B}'_{21}$. So in this case $\mathcal{B}'$ fails to separate all extensions of $f$ and $g$.

This is long. I originally thought this would work but it does not separate all extensions of $f$ and $g$ (see Nik's comment).

Consider the unital embedding $\mathcal A = M_2 \subset M_4$ given by $$A = \left[\begin{array}{cc}a&b\\c&d\end{array}\right] \mapsto A \otimes I_2 = \left[\begin{array}{cccc} a &&b\\&a&&b \\ c&&d\\&c&&d\end{array}\right]$$ and for $\gamma_n = -\sqrt \frac{n-1}{n} + \frac{i}{\sqrt{n}}\in\mathbb T$ let $$\mathcal B_n \equiv \left\{ \left[\begin{array}{cccc}a &&b \\ &a&&\gamma_nb\\c&&d\\&\bar{\gamma}_nc&&d\end{array}\right] : a,b,c,d\in \mathbb C\right\}$$ Then $\mathcal B$ is embedded in $\mathcal B_n$ which is a C$^*$-subalgebra of $M_4$. Finally, define $\mathcal B' \equiv \oplus_{n=1}^\infty \mathcal B_n \subset \oplus_{n=1}^\infty M_4 \equiv \mathcal A'$ with $A\in \mathcal A$ embedded as $\oplus_{n=1}^\infty (A\otimes I_2) \subset \mathcal A'$.

Claim 1: The distance from $P$ to $\mathcal B'$ is 1.

Let $\gamma_\infty = -1 = \lim_{n\rightarrow \infty} \gamma_n$ and define $\mathcal B_\infty$ in the same manner as above. What is the distance between $\mathcal B_\infty$ and $P\otimes I_2$? This is the same as calculating the minimum norm of $\left[\begin{array}{cc} 1+b&\\&1-b\end{array}\right], b\in\mathbb C$ which is 1 since $2 = |1 + b + 1 - b| \leq |1 + b| + |1 -b|$. Because of the definition of these matrix algebras it is straightforward to see that $\lim_{n\rightarrow \infty} d(\mathcal B_n, P\otimes I_2) = d(\mathcal B_\infty, P\otimes I_2) = 1$. Therefore, $$d(\mathcal B', \oplus_{n=1}^\infty (P\otimes I_2)) = 1$$

Claim 2: $\mathcal B'$ separates any two states on $\mathcal A'$ extending $f$ and $g$.

Let $v' = (v_1,v_2,v_3,v_4)\in \mathbb C^4$ such that $\langle (A\otimes I_2) v',v'\rangle = \langle Av,v\rangle$ for all $A\in M_2$. An easy calculation gives that $\|(v_1,v_2)\| = \frac{1}{\sqrt{2}}$ and $(v_1,v_2) = (v_3,v_4)$. The $w'$ case follows similarly except with $(w_1,w_2) = -(w_3,w_4)$.

Now for the general case, $\mathcal A'$ is naturally represented on the Hilbert space $\mathcal H = \oplus_{n=1}^\infty \mathbb C^4$. Let $v', w'\in \mathcal H$ be unit vectors such that $f'(C) = \langle Cv',v'\rangle$ and $g'(C) = \langle Cw',w'\rangle$ are states extending $f$ and $g$ respectively. From the previous paragraph they must have a particular form, namely $v' = (r_1,s_1,r_1,s_1, r_2,s_2,r_2,s_2,\cdots)$ and $w' = (t_1,u_1,-t_1,-u_1, t_2,u_2, -t_2,-u_2,\cdots)$.

Define $B_n = \left[\begin{array}{cccc}0&&1\\&0&&\gamma_n \\ \gamma_n&&0\\ &1&&0\end{array}\right] \in \mathcal B_n$ and let $B' = \oplus_{n=1}^\infty B_n \in \mathcal B'$. Now, $$Im\left\langle B_n\left(\begin{array}{c} r_n\\s_n\\r_n\\s_n\end{array}\right),\left(\begin{array}{c} r_n\\s_n\\r_n\\s_n\end{array}\right)\right\rangle = Im(1+\gamma_n)(\|r_n\|^2 + \|s_n\|^2) = \frac{1}{\sqrt{n}}(\|r_n\|^2 + \|s_n\|^2)$$ whereas $$Im\left\langle B_n\left(\begin{array}{c} t_n\\u_n\\-t_n\\-u_n\end{array}\right),\left(\begin{array}{c} t_n\\u_n\\-t_n\\-u_n\end{array}\right)\right\rangle = Im(1+\gamma_n)(-\|t_n\|^2 - \|u_n\|^2) = -\frac{1}{\sqrt{n}}(\|t_n\|^2 + \|u_n\|^2).$$ Hence, $$Im\langle B'v',v'\rangle = \sum_{n=1}^\infty \frac{1}{\sqrt{n}}(\|r_n\|^2+\|s_n\|^2) > 0$$ and $$Im\langle B'w',w'\rangle = -\sum_{n=1}^\infty \frac{1}{\sqrt{n}} (\|t_n\|^2 + \|u_n\|^2) < 0.$$ Therefore, $f'(B') \neq g'(B')$.