The following question is related to this, which was asked earlier on the forum. I am still a bit confused by the situation and decided to ask a separate question about it.
Let $C\subset B\subset A$ be an inclusion of $C^*$-algebras with $1_{A}=1_{B}$.
Assume that $C$ is a non-zero hereditary $C^*$-sub-algebra of $A$. Moreover, there is a conditional expectation $E_B:A\to B$.
Question: Is it true that every state on $B$ extends uniquely to a state on $A$?
Since, $C$ is a hereditary $C^*$-sub-algebra of $A$, every state on $C$ extends uniquely to a state on $A$. Now, given a state $\tau$ on $B$, $\tau_e^1$ and $\tau_e^2$ be any two possible extensions of $\tau$ to $A$. Now, $$\tau_e^1|_C=\tau_e^2|_C,$$ and hence, by the unique-extension property of $C$ inside $A$, they must agree. Moreover, $\tau\circ E_B$ is also one possible extension of $\tau$. So, this is the only possible extension.
Is this alright?
No, this is not true.
Consider$\def\CC{\mathbb C}$ $$ A=M_2(\CC),\qquad B=\Big\{\begin{bmatrix} a&0\\0&d\end{bmatrix}:\ a,d\in\CC\Big\},\qquad C=\Big\{\begin{bmatrix} a&0\\0&0\end{bmatrix}:\ a\in\CC\Big\}. $$ Then $C$ is hereditary (both in $A$ and in $B$) and $B\subset A$ with the same unit.
Consider the normalized trace on $B$, $$ \varphi\Big(\begin{bmatrix} a&0\\0&d\end{bmatrix}\Big)=\frac{a+d}2, $$ and the two extensions to $A$ $$ \psi_1\Big(\begin{bmatrix} a&b\\c&d\end{bmatrix}\Big)=\frac{a+d}2,\qquad\qquad\psi_2\Big(\begin{bmatrix} a&b\\c&d\end{bmatrix}\Big)=\frac{a+b+c+d}2. $$ Where your argument fails is that the restriction of a state to a subalgebra is not necessarily a state. So every state in $C$ admits a unique extension, but $B$ has states (as in the example) that do not restrict to states of $C$.