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 $x + Z_{\mathfrak{u}}(y)$.

We write the $\mathfrak{sl}_2$-module $\mathfrak{u}$ as $\bigoplus_{\lambda} \mathfrak{u}_{\lambda}$ where $\mathfrak{u}_{\lambda}$ is the isotypic part of highest weight $\lambda$. We also write $\mathfrak{u}_{\lambda}(i)$ to denote the vectors in $\mathfrak{u}_{\lambda}$ with $h$-weight $i$. Then it is clear that $Z_{\mathfrak{u}}(y) = \bigoplus_i \mathfrak{u}_i(-i)$ and hence by $\mathfrak{sl}_2$ representation theory, we get $\mathfrak{u} = Z_{\mathfrak{u}}(y) + [x,\mathfrak{u}]$.

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

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 $u'_0 \cdot x = x - [x,n'_0]$. 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 $n_0$. Let $c_1 = c_0 + c'_1$. Now we have $u'_0 \cdot (x+c_0+n_0) = x+c_1+n_1$. We repeat the argument again, observing that $n_1 = c'_2 + [x, n'_1]$ implies that $c'_2$ is at least as deep in the central series as $n_1$.

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.

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.

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 $\mathrm{Lie} M$ 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 $x + Z_{\mathfrak{u}}(y)$.

We write the $\mathfrak{sl}_2$-module $\mathfrak{u}$ as $\oplus_{\lambda} \mathfrak{u}_{\lambda}$ where $\mathfrak{u}_{\lambda}$ is the isotypic part of highest weight $\lambda$. We also write $\mathfrak{u}_{\lambda}(i)$ to denote the vectors in $\mathfrak{u}_{\lambda}$ with $h$-weight $i$. Then it is clear that $Z_{\mathfrak{u}}(y) = \oplus_i \mathfrak{u}_i(-i)$ and hence by $\mathfrak{sl}_2$ representation theory, we get $\mathfrak{u} = Z_{\mathfrak{u}}(y) + [x,\mathfrak{u}]$.

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

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 $u'_0 \cdot x = x - [x,n'_0]$. 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 $n_0$. Let $c_1 = c_0 + c'_1$. Now we have $u'_0 \cdot (x+c_0+n_0) = x+c_1+n_1$. We repeat the argument again, observing that $n_1 = c'_2 + [x, n'_1]$ implies that $c'_2$ is at least as deep in the central series as $n_1$.

Eventually, this must terminate, and at that point we have found some product $u'_k...u'_2u'_1u'_0$ that conjugates $x+n$ to $x+c_{k+1}$ as desired.

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 $x + Z_{\mathfrak{u}}(y)$.

We write the $\mathfrak{sl}_2$-module $\mathfrak{u}$ as $\bigoplus_{\lambda} \mathfrak{u}_{\lambda}$ where $\mathfrak{u}_{\lambda}$ is the isotypic part of highest weight $\lambda$. We also write $\mathfrak{u}_{\lambda}(i)$ to denote the vectors in $\mathfrak{u}_{\lambda}$ with $h$-weight $i$. Then it is clear that $Z_{\mathfrak{u}}(y) = \bigoplus_i \mathfrak{u}_i(-i)$ and hence by $\mathfrak{sl}_2$ representation theory, we get $\mathfrak{u} = Z_{\mathfrak{u}}(y) + [x,\mathfrak{u}]$.

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

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 $u'_0 \cdot x = x - [x,n'_0]$. 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 $n_0$. Let $c_1 = c_0 + c'_1$. Now we have $u'_0 \cdot (x+c_0+n_0) = x+c_1+n_1$. We repeat the argument again, observing that $n_1 = c'_2 + [x, n'_1]$ implies that $c'_2$ is at least as deep in the central series as $n_1$.

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.