Given is a unital $C^*$-algebra $A$ with state $\mathbb E$ and closed sub-$*$-algebra $B$ (unital, with the same identity). Think of $(A,\mathbb E)$ as a (possibly) non-commutative probability space.
How can one define (and prove the existence of) the "conditional expectation" w.r.t. $B$?
I am not quite sure what this is even supposed to be or how much extra structure may be needed for this, but at the very least it should be a positive, linear, unital map $\mathbb E_B : A \to B$ s.t. $\mathbb E_B(bab') = b\mathbb E_B(a)b'$ for $b,b'\in B$.
Furthermore, it should generalize the classical conditional expectation when $A = L_\infty(\Omega,\mathcal F,\mathbb P)$ and $B = L_\infty(\Omega,\mathcal G,\mathbb P)$, where $\mathcal G$ is a sub-$\sigma$-algebra of $\mathcal F$. The last property in the previous paragraph correspond to the property that if $X,Y$ are random variables and $Y$ is $\mathcal G$-measurable, then $\mathbb E(XY|\mathcal G) = Y\mathbb E(X|\mathcal G)$.
There's certainly no general recipe to construct a conditional expectation. There are a few known situations and techniques that give expectations on von Neumann algebras. There's certainly nothing general in the case of C$^*$-algebras.
And there is no guarantee an expectation exists: for a easy example, take $A=B(H)$, with $H$ infinite-dimensional, and $B=K(H)$, the compact operators. If you have an expectation $E:B(H)\to K(H)$, then $E$ preserves rank-one operators. Now let $T\in B(H)$, $\xi\in H$, and $P$ the projection onto $\mathbb C\xi$. Then $$ E(T)\xi=E(T)P\xi=E(TP)\xi=TP\xi=T\xi $$ (since $TP$ is finite-rank, $E(TP)=TP$. As $\xi$ is arbitrary, we conclude that $E(T)=T$, a contradiction since the range of $E$ is supposed to be $K(H)$. That is, no such expectation exists.