Let $A$ be a von Neumann algebra and $B$ is a strongly closed hereditary $C^*$-subalgebra of $A$. Prove that, there exists a unique projection $p$ in $B$ such that $B=pAp.$
$\underline{\text{Definition:}}$ A $C^*$-subalgebra $B$ of a $C^*$-algebra $A$ is called hereditary if for $a \in A$ and $b \in B$ the inequality $0\le a\le b$ implies $a \in B$.
My approach: Since $A$ is von Neumann algebra and $B$ is a strongly closed $C^*$-subalgebra of $A$, then $B$ is also a von Neumann algebra. Assuming $B \neq 0,$ it is unital and let the unit in $B$ be $p$. I think this is the required projection. But how can I show that $pAp=B?$ And what about the uniqueness? I have to use hereditary condition somewhere, but I failed.
Please help me to solve this. Thanks in advance.