= Solution
Take an environment initially in its vacuum state $|0\rangle_E$, with $|1\rangle_E$ recording an emitted photon. In atom-environment order, put $q=\sqrt{1-p}$ and $u=\sqrt p$. A <unitary dilation of amplitude damping> is defined by
$$
\begin{aligned}
U|00\rangle&=|00\rangle,& U|10\rangle&=q|10\rangle+u|01\rangle,\\
U|01\rangle&=q|01\rangle-u|10\rangle,& U|11\rangle&=|11\rangle.
\end{aligned}
$$
The middle two images are orthonormal, and they are orthogonal to the two fixed images. Thus this completion really is a <unitary operator>. For the occupied initial environmental subspace, it leaves a ground-state atom unchanged and splits an excited-state atom into no-emission and emission alternatives with probabilities $1-p$ and $p$.
Tracing out the environment after this unitary evolution gives
$$
\Phi(\rho)=\operatorname{Tr}_E\bigl[U(\rho\otimes|0\rangle\langle0|_E)U^\dagger\bigr].
$$
The environment <matrix> elements are $\langle0|_EU|0\rangle_E=A_0$ and $\langle1|_EU|0\rangle_E=A_1$, recovering the <Kraus operators> in part (a). More explicitly, the basis operators transform as
$$
\begin{aligned}
|0\rangle\langle0|&\longmapsto|0\rangle\langle0|,\\
|1\rangle\langle1|&\longmapsto(1-p)|1\rangle\langle1|+p|0\rangle\langle0|,\\
|0\rangle\langle1|&\longmapsto\sqrt{1-p}|0\rangle\langle1|,\\
|1\rangle\langle0|&\longmapsto\sqrt{1-p}|1\rangle\langle0|.
\end{aligned}
$$
Applying linearity to the atomic <density operator> therefore yields
$$
\boxed{\Phi(\rho)=\begin{pmatrix}
\rho_{00}+p\rho_{11}&\sqrt{1-p}\,\rho_{01}\\
\sqrt{1-p}\,\rho_{10}&(1-p)\rho_{11}
\end{pmatrix}.}
$$
The lost excited population goes to the ground state, the <trace> stays one, and coherences decrease by $\sqrt{1-p}$. Discarding the distinguishable environmental photon record is precisely what turns the joint unitary evolution into a noisy atomic <quantum channel>.
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