= Solution
Use the rotation-invariant distribution over pure initial states, namely the <uniform pure-qubit average> with area element $\sin\theta\,d\theta\,d\varphi/(4\pi)$. Reflection symmetry gives $\mathbb E[xz]=0$, and rotational symmetry together with $x^2+y^2+z^2=1$ gives $\mathbb E[y^2]=1/3$. Averaging the <squared quantum fidelity> from part (b) therefore yields
$$
\boxed{\overline{\langle\psi_d|\rho|\psi_d\rangle}=1-\frac1{2k}\quad(k\ge2).}
$$
If the single-number packet $k=1$ is included, its corrected average is
$$
\boxed{\overline{\langle\psi_d|\rho|\psi_d\rangle}=\frac7{12}\quad(k=1).}
$$
Uniform polar angle without the sine weight would not be the Haar-uniform average over all pure <qubit> states.
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