= Solution
Choose the <exchange-positive CP convention for neutral kaon mixing>, $CP|K^0\rangle=|\bar K^0\rangle$. Neglecting <CP violation>, the <kaon CP eigenstates> can then be written
$$
|K_S\rangle=\frac{|K^0\rangle+|\bar K^0\rangle}{\sqrt2},\qquad
|K_L\rangle=\frac{|K^0\rangle-|\bar K^0\rangle}{\sqrt2}.
$$
Writing $A=A(\bar D^0\to K^0\pi^0)$ and $B=A(\bar D^0\to\bar K^0\pi^0)$, projection on these coherent final states gives
$$
\boxed{A_S=\frac{A+B}{\sqrt2}=\frac{A}{\sqrt2}(1+B/A),\qquad
A_L=\frac{A-B}{\sqrt2}=\frac{A}{\sqrt2}(1-B/A).}
$$
This is the convention implicit in the printed sum. Changing the sign convention for the <neutral antikaon> changes both its coefficient in the <kaon CP eigenstates> and $B$; it cannot change either physical <decay rate>. The extremely small $K_S$–$K_L$ <mass> difference can be ignored in the two-body <phase space>.
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