In the proof of the BDF Theorem in the book Analytic K-Homology by Higson/Roe (chapter 7, page 187, I pasted the proof below) the authors claim that given an essentially normal operator $T$ there is a normal operator $T_{1}$ such that $PT_{1}P$ is equal to $T$ modulo compacts which is shown in Chapter 3. Looking through chapter 3 several times I wasn't able to find what the authors are referencing here.
Does anyone who knows the book could help me finding it?
The information is there in Chapters 2 and 3, but not in a way that one immediately sees the assertion as phrased in the paragraph you quote. It might be obvious to an expert (not my case), but it requires some digging otherwise. Here is what happens:
By Theorem 3.4.8, $\def\ext{\operatorname{Ext}}\ext(C(X))$ is a group.
Then by Proposition 2.7.3, $\ext(X)$ is a group.
Proposition 2.7.5 tells us that the zero of $\ext(X)$ is the class of split extensions.
In light of the above, Theorem 3.4.8 says that there exists an essentially normal operator $S$, with $\sigma_{\rm ess}(S)=X$, such that $[T]\oplus[S]=[0]$.
A split extension consists of a C$^*$-algebra $A\subset B(H)$ and a $*$-homomorphism $\rho:A\to C(X)$ such that there exists a $*$-homomorphism $\psi:C(X)\to B(H)$ such that $\rho\circ\psi=\text{id}_{C(X)}$.
Given a split extension, we can put $N=\psi(\text{id}_X)\in B(H)$. This is a normal operator with $\sigma(N)=X$. We can then choose $(C^*(N),\rho)$ as a split extension.
The equality $[T]\oplus[S]=[0]$ means that there exist a unitary $U$ and a compact operator $K$ such that $$ \begin{bmatrix} T&0\\0&S\end{bmatrix}=UNU^*+K. $$ Thus $T_1=UNU^*$ is normal, acting on a space that contains $H$, and $$ PT_1P=T-PKP, $$ with $PKP$ compact.