This is very easy to confirm when $n=2$: We can then also assume that $Pe_1=e_1$, so $$ P=\begin{pmatrix} 1 & a \\ 0 & 0 \end{pmatrix} , $$ and $\|P\|=\| 1-P\| = \sqrt{1+|a|^2}$.

In general, pick an $x$ with $\|x\|=1, \|Px\| = \|P\|$ and restrict $P$ to the invariant subspace $V$ spanned by $x,Px$. Then $\dim V=2$ and $P\not= 0,1$ also on $V$, unless we are in the trivial case $\|P\|=1$, so the first part shows that $(1-P)\bigr|_V=\|P\|$. Thus $\|1-P\|\ge \|P\|$ and then also $\|1-P\|=\| P\|$ by symmetry.

(This argument, very slightly modified, also works in general, when $\dim H=\infty$.)

This is very easy to confirm when $n=2$: We can then also assume that $Pe_1=e_1$, so $$ P=\begin{pmatrix} 1 & a \\ 0 & 0 \end{pmatrix} , $$ and $\|P\|=\| 1-P\| = \sqrt{1+|a|^2}$.

In general, pick an $x$ with $\|x\|=1, \|Px\| = \|P\|$ and restrict $P$ to the invariant subspace $V$ spanned by $x,Px$. Then $\dim V=2$ and $P\not= 0,1$ also on $V$, unless we are in the trivial case $\|P\|=1$, so the first part shows that $\|(1-P)\bigr|_V\|=\|P\|$. Thus $\|1-P\|\ge \|P\|$ and then also $\|1-P\|=\| P\|$ by symmetry.

(This argument, very slightly modified, also works in general, when $\dim H=\infty$.)