= Solution
Write $P_0=|\psi_0\rangle\langle\psi_0|$ and $Q_0=I-P_0$. <Frustration freeness> implies that every local <orthogonal projection> fixes $|\psi_0\rangle$ on both the left and right. Hence
$$
KP_0=P_0K=P_0,\qquad
K=P_0+Q_0KQ_0.
$$
The <spectral gap> makes $\operatorname{supp}H=Q_0\mathcal H$. The supplied <detectability lemma> gives
$$
q:=\left(1+\frac{\Delta}{2}\right)^{-1/3}<1,\qquad
\|Q_0KQ_0\|\leq q.
$$
Consequently
$$
\boxed{\|K^m-P_0\|\leq q^m=e^{-\alpha m},
\qquad
\alpha=\frac13\ln\left(1+\frac{\Delta}{2}\right).}
$$
No assumption that $K$ is Hermitian or normal is needed.
A stronger estimate will keep this same $\alpha$ when we count the two projection layers in (ii). On $Q_0\mathcal H$, put $P=P_{\rm odd}-P_0$ and $Q=P_{\rm even}-P_0$. They are <orthogonal projections>, with $\|PQ\|\leq q$. For $m\geq1$,
$$
(PQ)^m=PQ(QPQ)^{m-1},\qquad
\|QPQ\|=\|PQ\|^2,
$$
so the <powers of a product of two orthogonal projections> satisfy
$$
\boxed{\|K^m-P_0\|\leq q^{\,2m-1}.}
$$
This extra estimate uses the projection structure, rather than generic submultiplicativity alone.
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