= Solution
Label the atomic ground state $|0\rangle$ and excited state $|1\rangle$, with $0\leq p\leq1$. The process is the <amplitude damping channel>, with <Kraus operators>
$$
\boxed{A_0=|0\rangle\langle0|+\sqrt{1-p}\,|1\rangle\langle1|
=\begin{pmatrix}1&0\\0&\sqrt{1-p}\end{pmatrix},
\qquad
A_1=\sqrt p\,|0\rangle\langle1|
=\begin{pmatrix}0&\sqrt p\\0&0\end{pmatrix}.}
$$
They satisfy $A_0^\dagger A_0+A_1^\dagger A_1=I$, so the resulting <Kraus representation> is <trace> preserving.
The operator $A_1$ represents emission: it maps the excited state to $\sqrt p$ times the ground state and annihilates the ground state. Its probability for an initially excited atom is $p$. The operator $A_0$ represents no emission: it leaves the ground state unchanged and retains the excited amplitude with factor $\sqrt{1-p}$. This branch is not generally the identity operation; the absence of a photon itself changes the conditional relative amplitudes. For an arbitrary atomic <density operator>, the emission probability is $\operatorname{Tr}(A_1\rho A_1^\dagger)=p\rho_{11}$ and the no-emission probability is $1-p\rho_{11}$.
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