= Solution
Introduce all four physical <probability amplitudes>: $A_f,\bar A_f,A_{\bar f},\bar A_{\bar f}$. Under exchange of the <Flavor eigenstate>[flavour] and final-state labels, define
$$
\boxed{\bar\lambda=\frac pq\frac{A_{\bar f}}{\bar A_{\bar f}}.}
$$
This <CP-conjugate decay-amplitude ratio> gives the conjugate-channel rates
$$
\boxed{\begin{aligned}
\Gamma(\bar K(t)\to\bar f)&=|\bar A_{\bar f}|^2\left[|g_+|^2+|\bar\lambda|^2|g_-|^2+2\operatorname{Re}(\bar\lambda g_-g_+^*)\right],\\
\Gamma(K(t)\to\bar f)&=|A_{\bar f}|^2\left[|g_+|^2+|\bar\lambda^{-1}|^2|g_-|^2+2\operatorname{Re}(\bar\lambda^{-1}g_-g_+^*)\right].
\end{aligned}}
$$
<CP> is a <Unitary operator>[unitary] discrete transformation, so exchanging state labels does not mean complex-conjugating every strong <phase> by fiat. Nor should one assume that CP-conjugate <probability amplitudes> are equal while studying <CP violation>.
The equality $\bar\lambda=\lambda^{-1}$ holds when
$$
\boxed{\frac{A_{\bar f}}{\bar A_{\bar f}}=\frac{A_f}{\bar A_f},}
$$
in particular when $\bar f$ and $f$ are the same <CP eigenstate> ray, with a consistent state <phase>. It is not automatic for two distinct CP-conjugate final states. For example, $p=q$, $A_f=\bar A_{\bar f}=2$ and $\bar A_f=A_{\bar f}=1$ conserve the paired decay <probability amplitudes> but give $\lambda=\bar\lambda=1/2$, not inverse values.
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