This is about Exercise 7.1.3. in Brown and Ozawa. They say that $A$ is quasidiagonal if and only if there exists an injective $*$-hom. $$ A \to \frac{\prod_n M_n(\mathbb C) }{\sum_n M_n(\mathbb C)} $$ which admits a c.p.c. splitting $A \to \prod_n M_n(\mathbb C)$.
I do not understand what it means to be a splitting here. Also I guess they are talking about separable $C^*$-algebras here, since they index over the natural numbers ?
I think the splitting means to have maps $$ A \xrightarrow{\ \ \ \ \alpha \ \ \ \ } {\prod_n M_n(\mathbb C) } \xrightarrow{\ \ \ \ \beta \ \ \ \ } \frac{\prod_n M_n(\mathbb C) }{\sum_n M_n(\mathbb C)}, $$ where $\alpha$ is cpc, $\beta$ is the quotient map, and $\beta\circ\alpha$ is equal to the original injective $*$-homomorphism.