Here is what I know. Assume $M\cong G/K$ is an irreducible hermitian symmetric space. Denote the Lie-algebra of $K$ by $\mathfrak{t}$. Proposition 1.2. chapter 3 in
[Wienhard - Bounded cohomology and geometry](https://arxiv.org/abs/math/0501258)
says that for a symmetric space, being hermitian is equivalent to the existence of a non-zero element $Z$ in the center of $\mathfrak{t}$. In particular $K\subset \operatorname{Stab}\_G(Z)$ is a subset of the stabilizer of $Z$ under the coadjoint action. If one could show that $K=\operatorname{Stab}\_G(Z)$, then $M=G/K$ can be identified with the coadjoint orbit $\mathcal{O}\_Z\cong G/\operatorname{Stab}\_G(Z)$.



Can all hermitian symmetric spaces be realised as coadjoint orbits?
This is true. One can use a few facts from Helgason's *Differential Geometry, Lie Groups, and Symmetric Spaces* to show that, indeed, $K = \mathrm{Stab}\_G(Z)$.


Since $M=G/K$ is an irreducible Hermitian symmetric space, one can apply Theorem 6.1 from Chapter VIII of Helgason. The main point is that one can regard $G$ as a subgroup of $\mathrm{Aut}(\frak{g})$ with $K\subset G$ connected, and the circle $\mathrm{exp}(\mathbb{R}\cdot Z)\simeq S^1$ is the center of $K$. (See Helgason's Proposition 6.2.) Moreover, as Helgason explains, there will be a real number $t$ such that the element $z = \exp(tZ)$ is an element of order $2$ with the property that the $G$-centralizer of $z\in K$ is equal to $K$. Since the $G$-stabilizer of $Z$ is clearly contained in the $G$-centralizer of $z$, it follows that the $G$-stabilizer of $Z$ must equal $K$ as well.


