The proof in Murphy's book (chapter on Von Neumann algebras) of the fact that a *-subalgebra $A$ of $B(\mathcal{H})$ is strongly dense in $A''$.
The proof procedes as follows: Take a $x\in \mathcal{H}$ and construct a closed subspace of $\mathcal{H}$ namely $$\mathcal{K}:=\overline{\{v(x) | v\in A\}}^{\|.\|}$$
Then define $p$ the projection of $\mathcal{H}$ onto $\mathcal{K}$.
Then the claim is that $p\in A'$. Why is this so?
Wouldn't this mean that $\forall x \in \mathcal{H}$ we have that $$p(v(x))=v(p(x)).$$ But $p(x)$ might be zero whereas $v(x)$ might not be. Wouldn't be a contradiction? Or is this impossible. In either case why would it be true that $p\in A'$?
Note that $pv(x) = v(x)$ for all $x\in H$ and $v\in A$ you get $pv = v$. In particular since $A$ is $*$-closed you have for all $v\in A$ that $vp = (p^*v^*)^*=(pv^*)^*=v$, and so $vp=v=pv$.