This is really an extended comment, but, because it's too long to put into a comment box and because it may help answer some of the OP's questions, I'm putting it here.

If you endow $\mathrm{SU}(n)$ with its usual bi-invariant measure $\mathrm{d}\mu$ normalized to have total volume $1$ (aka Haar measure), then we know that, for any $B\in{\frak{su}}(n)$, we have $$ \int_{U\in\mathrm{SU}(n)} \mathrm{Ad}_U(B)\ \mathrm{d}\mu = 0. $$ This implies that, for any $A,B\in {\frak{su}}(n)$ the average value of the function $f:\mathrm{SU}(n)\to\mathbb{R}$ defined by $f(U) = K\bigl(A,\mathrm{Ad}_U(B)\bigr)$ over $\mathrm{SU}(n)$ is also zero. Now, because the linear span of the $\mathrm{Ad}$-orbit of $B$ must be an $\mathrm{Ad}$-invariant subspace, it is either the zero subspace or all of ${\frak{su}}(n)$, so the only way that $f$ could vanish identically would be for either $A$ or $B$ to be zero.

In particular, it follows that, when $A$ and $B$ are both nonzero, the subset $S_{A,B} = f^{-1}(0)\subset \mathrm{SU}(n)$ divides its complement in $\mathrm{SU}(n)$ into two nonempty open sets and hence, since it is defined algebraically, it must be, at most places, a smooth hypersurface, i.e., it must have codimension $1$, at least where it is smooth.

Now, there are no subgroups of $\mathrm{SU}(n)$ that have codimension $1$, so $S_{A,B}$ cannot ever be a subgroup of $\mathrm{SU}(n)$, when both $A$ and $B$ are nonzero.

Of course, when $K(A,B)=0$, as I already mentioned, $S_{A,B}$ can contain subgroups, in particular, the subgroups $P$ and $Q$, which are the $\mathrm{Ad}$-stabilizers of $A$ and $B$, respectively. Most of the time, though, these will be maximal subgroups of $\mathrm{SU}(n)$ that lie in $S_{A,B}$ because, generically, $P$ and $Q$ will be maximal tori, and hence there will only be a finite number of connected Lie subgroups that lie between, say $P$ and $\mathrm{SU}(n)$ itself. When $B$ is generically chosen with respect to $A$, none of these properly sandwiched subgroups will lie in $S_{A,B}$, so $P$ will be maximal. A similar argument applies to $Q$.

This is really an extended comment, but, because it's too long to put into a comment box and because it may help answer some of the OP's questions, I'm putting it here.

If one endows $\mathrm{SU}(n)$ with its usual bi-invariant measure $\mathrm{d}\mu$ normalized to have total volume $1$ (aka Haar measure), then one knows that, for any $B\in{\frak{su}}(n)$, $$ \int_{U\in\mathrm{SU}(n)} \mathrm{Ad}_U(B)\ \mathrm{d}\mu = 0. $$ (The integral has to be an element of ${\frak{su}}(n)$ that is $\mathrm{Ad}$-invariant, so it must be zero.)

This implies that, for any $A,B\in {\frak{su}}(n)$ the average value of the function $f:\mathrm{SU}(n)\to\mathbb{R}$ defined by $f(U) = K\bigl(A,\mathrm{Ad}_U(B)\bigr)$ over $\mathrm{SU}(n)$ is also zero. Now, because the linear span of the $\mathrm{Ad}$-orbit of $B$ must be an $\mathrm{Ad}$-invariant subspace, it is either the zero subspace or all of ${\frak{su}}(n)$, so the only way that $f$ could vanish identically would be for either $A$ or $B$ to be zero.

In particular, it follows that, when $A$ and $B$ are both nonzero, the subset $S_{A,B} = f^{-1}(0)\subset \mathrm{SU}(n)$ divides its complement in $\mathrm{SU}(n)$ into two nonempty open sets and hence, since it is defined algebraically, it must be, at most places, a smooth hypersurface, i.e., it must have codimension $1$, at least where it is smooth. (Because $f$ is real-analytic and $\mathrm{SU}(n)$ is connected, its zero locus cannot contain any nonempty open set.)

Now, there are no subgroups of $\mathrm{SU}(n)$ that have codimension $1$, so $S_{A,B}$ cannot ever be a subgroup of $\mathrm{SU}(n)$ when both $A$ and $B$ are nonzero.

Of course, when $K(A,B)=0$, as I already mentioned, $S_{A,B}$ can contain subgroups, in particular, the subgroups $P$ and $Q$, which are the $\mathrm{Ad}$-stabilizers of $A$ and $B$, respectively. Most of the time, though, these will be maximal subgroups of $\mathrm{SU}(n)$ that lie in $S_{A,B}$ because, generically, $P$ and $Q$ will be maximal tori, and hence there will only be a finite number of connected Lie subgroups that lie between, say $P$ and $\mathrm{SU}(n)$ itself. When $B$ is generically chosen with respect to $A$, none of these properly sandwiched subgroups will lie in $S_{A,B}$, so $P$ will be maximal. A similar argument applies to $Q$.