The Harish-Chandra projection computes the highest-weight scalar

By Central elements lie in the zero-weight subspace of $U(\mathfrak{g})$, $z$ lies in $U(\mathfrak{g}{)}_0$. Fix the PBW order ${\mathfrak{n}}^{-},\mathfrak{h},{\mathfrak{n}}^{+}$ used in The Harish-Chandra projection. Every PBW monomial is an $\mathfrak{h}$-weight vector, with weight equal to the sum of its root weights. In the expansion of the zero-weight element $z$, only zero-weight monomials occur.

A zero-weight ordered monomial with a nontrivial ${\mathfrak{n}}^{-}$ factor must also have a nontrivial ${\mathfrak{n}}^{+}$ factor: without the latter its weight is a nonzero negative sum of positive roots. Thus every zero-weight PBW monomial outside $U(\mathfrak{h})$ has a rightmost factor in ${\mathfrak{n}}^{+}$ and annihilates $v$. Consequently $zv=\mathrm{pr}(z)v$. Merely having a left ${\mathfrak{n}}^{-}$ factor would not imply annihilation; the zero-weight restriction supplies the needed right factor.

The element $\mathrm{pr}(z)$ lies in $U(\mathfrak{h})$, so it acts on $v$ by the scalar obtained from the polynomial $\mathrm{pr}(z)$ by evaluation at the weight $\lambda$. Combining this with step 2.1 gives $zv=\mathrm{pr}(z)(\lambda )v$.