What is $\sigma_{\mathfrak{M}_{p}}(p\cdot a\cdot p)$ when $p=\chi_{S}(a)$ for closed $S\subseteq \sigma(a)$? Let $\mathfrak{M}$ be a von-Neumann algebra and $a\in\mathfrak{M}$ a self-adjoint element. Let $S\subseteq\sigma(a)$ be a non-empty, closed subset. Let $p=\chi_{S}(a)$, which is a non-zero, with-$a$-commuting projection.
Question. Is $\sigma_{\mathfrak{M}_{p}}(pap)=S$? If not, what then?
So far, I can show $\overline{S^{o}} \subseteq \sigma_{\mathfrak{M}_{p}}(pap) \subseteq S \subseteq \sigma_{\mathfrak{M}_{p}}(pap)\cup\sigma_{\mathfrak{M}_{1-p}}((1-p)a(1-p))$.
(That $\overline{S^{o}} \subseteq \sigma_{\mathfrak{M}_{p}}(pap)$, does not give much information, since $S\setminus S^{o}$ can be topologically quite large — i. e. non-meager.)

You can always "lose" zero; for instance say $a=p$, $S=\{0,1\}=\sigma(p)$. Then $p=\chi_S(p)$ and $\sigma_p(p)=1$ ($p$ is invertible in $p\mathfrak M p$).
You will also always lose the part of the spectrum that comes from the orthogonal complement of $p$. For example, let 
$$
a=\begin{bmatrix}1&0&0\\0&2&0\\0&0&3\end{bmatrix},\ \ p=\begin{bmatrix}1&0&0\\0&0&0\\0&0&0\end{bmatrix}, \ \ S=\{2,3\}.
$$
Then $S\subset\sigma(a)=\{1,2,3\}$, while $\sigma_p(ap)=\{1\}$.
