Suppose there exists a conditional expectation $E$ from the von Neumann algebra $M$ to its subalgebra $N$, the following inequality holds:
$E(a)^*E(a)\leq E(a^*a)$ for all $a\in M$.
Under what condition can make the equality $E(a)^*E(a)= E(a^*a)$ establish. Does there exist a sufficient and necessary to characterize the equality $E(a)^*E(a)= E(a^*a), \forall a\in M$.
That's the well-known Kadison Schwarz Inequality. It holds for unital completely positive maps (in reality, for unital $2$-positive maps). Below are a couple proofs.
If $\phi:M\to N$ is a completely positive map, the by Stinespring's Dilation Theorem you can write $\phi(x)=V^*\pi(x) V$ for a representation $\pi$ and an operator $V$. Since $\|V^*V\|=\|\phi(1)\|\leq\|\phi\|$ (this is actually an equality), $$ \phi(a)^*\phi(a)=V^*\pi(a^*)VV^*\pi(a)V\leq\|\phi\|\,V^*\pi(a^*)\pi(a)V=\|\phi\|\,V^*\pi(a^*a)V=\|\phi\|\,\phi(a^*a). $$
A second argument requires only $2$-positivity. The block-matrix $$ \begin{bmatrix} 1&a\\ a^*&a^*a\end{bmatrix} $$ is positive. Then $$ 0\leq\phi^{(2)}\Big(\begin{bmatrix} 1&a\\ a^*&a^*a\end{bmatrix}\Big) =\begin{bmatrix} \phi(1)&\phi(a)\\ \phi(a)^*&\phi(a^*a)\end{bmatrix} \leq\begin{bmatrix} \|\phi(1)\|&\phi(a)\\ \phi(a)^*&\phi(a^*a)\end{bmatrix}. $$ The positivity of this last matrix is equivalent to $\phi(a)^*\phi(a)\leq\|\phi(1)\|\,\phi(a^*a)$.