I thank Emile Okada for suggesting the following argument that there is a unique $P$-orbit in $q^{-1}(O) \cap p^{-1}(O_G)$ (all mistakes are due to me, however). The argument is inspired by Lemma 5.2.1 of

*DeBacker, Stephen*, **Parametrizing nilpotent orbits via Bruhat–Tits theory**, Ann. Math. (2) 156, No. 1, 295–332 (2002). ZBL1015.20033,
which appears to follow an argument of Waldspurger.

Let $x$ be a representative of $O$, with $q^{-1}(x) = x+\mathfrak{u}$. Each $P$ orbit in $q^{-1}(O)$ intersects $q^{-1}(x)$ and $q^{-1}(x)$ is normalized by $Z_M(x)U$. Hence, we are reduced to studying the $Z_M(x)U$-orbits in $q^{-1}(x)$.

We now extend $x$ to an $\mathfrak{sl}_2$-triple $(x,h,y)$ in $\operatorname{Lie} M$ and claim that any $Z_M(x)U$-orbit in $q^{-1}(x)$ will have a representative in $x + Z_{\mathfrak{u}}(y)$. If the claim is true, then we are done because $O_G$ intersects the Slodowy slice $x + Z_{\mathfrak{g}}(y)$ at just the point $x$.

To prove the claim, it suffices to show that $U \cdot (x + Z_{\mathfrak{u}}(y)) = x + \mathfrak{u}$. The inclusion of the left into the right is trivial, so we need only show the reverse inclusion. Let $x+n \in x + \mathfrak{u}$. We must find $u \in U$ conjugating $x+n$ into $x + Z_{\mathfrak{u}}(y)$.

Let $\mathfrak{u}=\mathfrak{u}_0 \supset \mathfrak{u}_1 \supset \dots \supset 0$ be the lower central series for the nilpotent Lie algebra $\mathfrak{u}$. We write the $\mathfrak{sl}_2$-module $\mathfrak{u}_i$ as $\bigoplus_{\lambda} \mathfrak{u}_{i, \lambda}$ where $\mathfrak{u}_{i, \lambda}$ is the isotypic part of $\mathfrak{u}_i$ of highest weight $\lambda$. We also write $\mathfrak{u}_{i, \lambda}(j)$ to denote the vectors in $\mathfrak{u}_{i, \lambda}$ with $h$-weight $j$. Then by $\mathfrak{sl}_2$ representation theory, we get that $Z_{\mathfrak{u}_i}(y) = \bigoplus_j \mathfrak{u}_{i, j}(-j)$ and hence $\mathfrak{u}_i = Z_{\mathfrak{u}_i}(y) + [x,\mathfrak{u}_i]$.

By a simple inductive argument on the lower central series of $\mathfrak{u}_i$, for any $n' \in \mathfrak{u}_i$, there exists $u' \in \mathrm{exp}(\mathfrak{u}_i)$ such that $u' \cdot x = x - [x,n']$ .

Now, let $c_0=0, n_0=n$. Write $n_0 = c'_1 + [x, n'_0]$ for $c'_1 \in Z_{\mathfrak{u}}(y)$. Let $u'_0$ be such that $u'_0 \cdot x = x - [x,n'_0]$. By the previous two paragraphs, $n'_0 , c'_1, u'_0 -1$ are as deep in the central series of $\mathfrak{u}$ as $n_0$.

Then, $u'_0 \cdot (x+ c_0 + n_0)= x + c_0 + c'_1 + n_1$ for some $n_1$ that is deeper in the lower central series than $n_0$. This is because $u'_0 \cdot (c_0 + n_0) = c_0 + n_0 + T$ where $T$ consists of terms involving lie brackets between $u'_0 -1$ and $c_0 + n_0$. Let $c_1 = c_0 + c'_1$. Now we have $u'_0 \cdot (x+c_0+n_0) = x+c_1+n_1$.

We now iterate the argument noting that eventually this must terminate, and at that point we have found some product $u'_k\dotsm u'_2u'_1u'_0$ that conjugates $x+n$ to $x+c_{k+1}$ as desired.