The normalized adjoint Casimir is a positive scalar

The adjoint representation is irreducible: an invariant subspace $U\subseteq \mathfrak{g}$ satisfies $[\mathfrak{g},U]\subseteq U$, so it is an ideal, and [F3] leaves only $0$ and $\mathfrak{g}$. By [F2], $T=\mathrm{ad}({\mathrm{\Omega }}_B)$ commutes with every $\mathrm{ad}(x)$. Write $n={\dim }_{\mathbb{C}}\mathfrak{g}>0$ by [F3], choose one finite basis, and let $A$ represent $T$. By [F7], $p(z)=\det (z{I}_n-A)$ is monic of degree $n$, hence is a nonconstant complex polynomial. By [F6] it has a root ${\gamma }_B\in \mathbb{C}$. By [F7], $\ker (T-{\gamma }_{B}I)\mathrm{\ne }0$. Since $T$ commutes with every $\mathrm{ad}(x)$, this eigenspace is invariant under $\mathfrak{g}$, hence is all of $\mathfrak{g}$. Therefore $T={\gamma }_B{\mathrm{id}}_{\mathfrak{g}}$, and this scalar is unique.

Choose a $Q=-B{\mathrm{\mid }}_{\mathfrak{k}}$-orthonormal basis $X_i$ of $\mathfrak{k}$. By [F4] and [F5], $Q$ is positive definite and the $B$-dual basis is $-X_i$, so ${\mathrm{\Omega }}_B=-{\sum }_i{X}_i^2$. Invariance of $Q$ makes each $\mathrm{ad}(X_i)$ skew-adjoint, and therefore $Q(TX,X)=Q\text{\hspace{0.17em}⁣}(-{\sum }_i[X_i,[X_i,X]],X)={\sum }_{i}Q([X_i,X],[X_i,X])\mathrm{.}$ For $X\mathrm{\ne }0$ this is positive: if it were zero, each $[X_i,X]$ would vanish, so $X$ would be central in $\mathfrak{k}$ and hence in $\mathfrak{g}$, contrary to [F3]. Since $T={\gamma }_{B}I$, dividing by $Q(X,X)>0$ shows that ${\gamma }_B$ is a positive real number.