= Solution
The state in part (b) is symmetric in the receiving <qubits>. Single out Bob using the <Dicke state> recursion. For $k\geq1$, the two <vectors> are
$$
\begin{aligned}
|D_k^N\rangle&=\sqrt{\frac{k+1}{N}}|0\rangle|D_k^{N-1}\rangle+\sqrt{\frac{k}{N}}|1\rangle|D_{k-1}^{N-1}\rangle,\\
|D_{k+1}^N\rangle&=\sqrt{\frac{k}{N}}|0\rangle|D_{k+1}^{N-1}\rangle+\sqrt{\frac{k+1}{N}}|1\rangle|D_k^{N-1}\rangle.
\end{aligned}
$$
Distinct excitation counts give <orthogonal> states of the other recipients. Taking their <partial trace>, the only surviving off-diagonal term comes from the common rest <vector> $|D_k^{N-1}\rangle$. The <one-qubit reduction of Dicke-state superpositions> is consequently
$$
\boxed{\rho=\frac1N\begin{pmatrix}k+|\alpha|^2&(k+1)\alpha\beta^*\\(k+1)\alpha^*\beta&k+|\beta|^2\end{pmatrix},\qquad k=\frac{N-1}{2}.}
$$
Put $p=|\alpha|^2$ and $q=|\beta|^2$, so $p+q=1$. The overlap with the input is
$$
\begin{aligned}
\langle\psi|\rho|\psi\rangle&=\frac{p(k+p)+q(k+q)+2(k+1)pq}{N}\\
&=\frac{k+p^2+q^2+2(k+1)pq}{N}=\frac{k+1+2kpq}{N}\\
&=\boxed{\frac{N+1+2|\alpha|^2|\beta|^2(N-1)}{2N}}.
\end{aligned}
$$
For $N=1$ there is no system to trace out, and the same <matrix> is $|\psi\rangle\langle\psi|$, with overlap one. For every $N>1$ the overlap is below one, even at its maximum $p=q=1/2$, so these are imperfect copies.
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