You're most welcome.
The rest of the expansions seem right to me.
As to proof 4.3, I see what you're doing there, and it's correct too, AFAICS. The only glitch is for these kind of proofs is that it's perhaps best to drop Dirac's notation, because it kind of stands in the way of distinguishing the vector, the operator, and the action of the inner product more clearly, so you wouldn't have to use the --somewhat awkward, IMO-- double parenthesis on your last line.
So, for example, I would write something like,
for every \( w \), \( v \) in \( \mathscr{H} \) (the Hilbert space of states),
\[ \left( w, P\right) = \left( P^{\dagger}w,v \right) \]
(That is just a definition of \( P^{\dagger} \), of course)
I would also write,
\[ v=v_{perp}+v_{parallel} \]
etc, with,
\[ Pv_{perp}=0 \]
\[ Pv_{parallel}=v_{parallel} \]
for an arbitrary vector \( v \).
And then, as you say,
\[ \left( w, Pv \right) = \left( w,w_{\parallel} \right) = \left( w_{\parallel}, v_{\parallel} \right) = \left( w_{\parallel}, v \right) = \left( Pw, v \right) \]
So indeed \( P^{\dagger}=P.
To me, it's a bit more transparent with this notation, but I understood what you meant, and if you think about it we're saying the same thing.
The devil is in the details, as they say. In infinite dimension one would have to be much more careful than this, but I don't have the chops for it. 😊
Ok. Something got messed up in the LaTeX rendering, I'm afraid... I hope you can see what it is.
I always have to be very careful that my text is not interpreted as rich text at some point (eg, when the editor refreshes) so the rendering is messed up.
It might have to do with the software at my end. I dunno.