Because the definition requires cutting it with central projections. And $B(H)$ is a factor, so the only nonzero projection is the identity itself.
What makes a projection not properly infinite is that you cannot do halving. For instance, consider $ M=\mathbb C\oplus B(H)$, and take the identity $1\oplus I$. You cannot halve this projection, because you would need to halve $1\in\mathbb C$.
You never mention how you define $M\rtimes_\alpha G$. In the direct sum picture, one defines, using the matrix units $\{E_{gh}:\ g,h\in G\}$,
$$
M\rtimes_\alpha G=\Bigg[\Big\{\sum_{g\in G}\alpha_g(x)\,E_{gg}:\ x\in M\Big\}\cup\Big\{\sum_{h\in G}E_{h,g^{-1}h}:\ g\in G\Big\}\Bigg]'',
$$
where $\hat x=\sum_{g\in G}\alpha_g(x)\,E_{gg}$ is how we see $M$ inside $M\rtimes_\alpha G$. The elements $u_g=\sum_{h\in G}E_{h,g^{-1}h}$ are unitaries, and the algebra $\hat M$ of elements $\sum_{g\in G}\alpha_g(x)\,E_{gg}$ is invariant under conjugation by each $u_g$.
The expectation is indeed "compression to the diagonal" in a block sense. And the "diagonal" is $\hat M$. In this notation, the projections are $E_{gg}$. Then
$$
\sum_{g\in G} E_{gg}\Big(\sum_{s\in G}\alpha_s(x)\,E_{ss}+\sum_{r\in G\setminus\{e\}} c_{r}\sum_{t\in G}E_{r^{-1}t,t}\Big)E_{gg}=\sum_{s\in G}\alpha_s(x)\,E_{ss}\in\hat M.
$$
To see that $\phi$ is normal, suppose that $T_\alpha\nearrow T$, selfadjoints. Given a vector $x=\sum_g x_g$, where $x_g=P_g x$,
\begin{align}
\|\phi(T-T_\alpha)x\|^2
&=\Big\|\sum_gP_g(T-T_\alpha)P_g\sum_hx_h\Big\|^2\\[0.2cm]
&=\Big\|\sum_gP_g(T-T_\alpha)P_g x_g\Big\|^2\\[0.2cm]
&=\sum_g\|P_g(T-T_\alpha) x_g\|^2\\[0.2cm]
&\leq\sum_g\|(T-T_\alpha) x_g\|^2\\[0.2cm]
\end{align}
Now because $T-T_\alpha$ is uniformly bounded and $\sum_g\|x_g\|^2<\infty$, given $\varepsilon>0$ there exists $F\subset G$, finite, such that $\sum_{g\in G\setminus F}\|(T-T_\alpha)x_g\|^2<\varepsilon$. Then
$$
\limsup_\alpha\|\phi(T-T_\alpha)x\|^2
\leq \varepsilon+\limsup_\alpha\sum_{g\in F}\|(T-T_\alpha)x_g\|^2=\varepsilon.
$$
This shows that $\phi(T_\alpha)\to\phi(T)$ sot, and so $\phi$ is normal.
Best Answer
If $A$ is positive and invertible, let $\lambda=\min\sigma(A)$. Then $A\geq\lambda I$. Thus $$ pAp\geq \lambda p, $$ and so $\sigma(pAp)\subset [\lambda,\infty)$ and $pAp$ is invertible in $pB(H)p$.
But the equality $(pAp)^{-1}=pA^{-1}p$ does not usually hold. For instance take $H=\mathbb C^2$, and $$ A=\begin{bmatrix} 2&1\\1&1\end{bmatrix},\ \ p=\begin{bmatrix} 1&0\\0&0\end{bmatrix}. $$ Then, as $A^{-1}=\begin{bmatrix} 1&-1\\-1&2\end{bmatrix}$, we have $$ pAp=2p\ne p=pA^{-1}p. $$