= Solution
Convergence in the <Hilbert-Schmidt norm> implies
$$
\lVert L_i^p\rVert_{\mathrm{HS}}^2
\longrightarrow\lVert L_i\rVert_{\mathrm{HS}}^2.
$$
Products of two Hilbert-Schmidt operators are <trace-class operators>, and the <Schatten norm Hölder inequality> gives
$$
\begin{aligned}
\lVert(L_2^p)^*L_1^p-L_2^*L_1\rVert_1
&\leq
\lVert L_2^p-L_2\rVert_{\mathrm{HS}}\lVert L_1^p\rVert_{\mathrm{HS}}\\
&\quad+\lVert L_2\rVert_{\mathrm{HS}}\lVert L_1^p-L_1\rVert_{\mathrm{HS}}
\longrightarrow0.
\end{aligned}
$$
By <trace duality>, every unitary $R$ satisfies
$$
\left|\operatorname{tr}\!\left(R^*\{(L_2^p)^*L_1^p-L_2^*L_1\}\right)\right|
\leq\lVert(L_2^p)^*L_1^p-L_2^*L_1\rVert_1.
$$
Taking the supremum over $R$ shows that the supremum terms in the two Procrustes formulas converge. Combining this with convergence of the squared norms proves
$$
d_P(C_1^p,C_2^p)^2\longrightarrow d_P(C_1,C_2)^2.
$$
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