I assume that you want $A$ to be a $\ast$-subalgebra.

If it so, then $A$ has to be a commutative von Neumann algebra, hence isomorphic to $L^{\infty}(X,\mu)$ for some measure space $X$. Every Banach space of this form is an injective Banach space (see a nice proof by Bill Johnson), i.e. every operator from a Banach space $V \subset W$ to $L^{\infty}(X,\mu)$ extends to an operator from $W$ to $L^{\infty}(X, \mu)$ with the same norm. To prove that there is a projection of norm one onto $A$, just extend the identity map $\mbox{Id}:A \to A$ to an operator $\widetilde{\mbox{Id}}:B(H) \to A$.

I assume that you want $A$ to be a $\ast$-subalgebra.

If it so, then $A$ has to be a commutative von Neumann algebra, hence isomorphic to $L^{\infty}(X,\mu)$ for some measure space $X$. Every Banach space of this form is an injective Banach space (see a nice proof by Bill Johnson), i.e. every operator from a Banach space $V \subset W$ to $L^{\infty}(X,\mu)$ extends to an operator from $W$ to $L^{\infty}(X, \mu)$ with the same norm. To prove that there is a projection of norm one onto $A$, just extend the identity map $\mbox{Id}:A \to A$ to an operator $\widetilde{\mbox{Id}}:B(H) \to A$.