Consider following theorem from Murphy's '$C^*$-algebras and operator theory'
The proof says that if $p=1$, then the assertion that the map $H \to \bigoplus_\lambda p_\lambda(H)$ is a unitary is clear.
I don't see why this is true though. To show it is a unitary, it suffices to show that it is isometric and surjective. I can see it is isometric, but don't see why it should be surjective.
I tried the following:
Let $(p_\lambda(x_\lambda))_\lambda \in \bigoplus_\lambda p_\lambda(H)$. I guess we must take something like $x= \sum_\lambda x_\lambda$ and show that this still gets mapped to what we want?
The map is obviously surjective.
Let $y\in \oplus_{\lambda\in \Lambda} p_{\lambda}(H)$ and define $x=\sum_{\lambda} y_{\lambda}$. We just need to argue that $p_{\lambda}(x)=y_{\lambda}$. Since, the $p_{\lambda}$ are orthogonal, we see that $p_{\lambda}(y_{\lambda'})=0$ for $\lambda\neq \lambda'$ and thus, by continuity
$$ p_{\lambda}(x)=\sum_{\lambda'}p_{\lambda}(y_{\lambda'})=p_{\lambda}(y_{\lambda})=y_{\lambda} $$ since $p_{\lambda}$ is a projection. This proves the desired.