Thank you. With this approach we are able to answer the question in affirmative.

But it holds a+b=1, so b=f(a).

I am interested in the situation of general $C^*$-algebras, and some "nontrivial" correspondence between $a$ and $b$. Since the correspondence in $B(H)$ is quite "strong", I expected that something nonobvious can be said also in general $C^*$-algebras.

Since the condition trivially holds if A is commutative, we exclude that case. In the noncommutative case, does it hold the same correspondence as in $B(H)$?

Yes, I am assuming that a and b are self-adjoint. Sorry for the mistake and thank you for pointing it out.

I cannot find the question in Halmos books. There were some other useful information. Thank you. For case $n=2$, might Mathematica be able to compute this?

But $x^∗ax$ is also hermitian matrix if $a$ is. So $x^∗Px⊂P$, and $x^∗M_n(\mathbb{C})x=M_n(\mathbb{C})$ iff $x$ is invertible. So $x^∗Px$ either does not satisfy (iv) or equals $P$.