= Solution
Within each fixed <unitary irreducible representation>, choose the selected <right singular vectors> to be <orthonormal>. Their corresponding <left singular vectors> are also <orthonormal>, even when a <singular value> is repeated: choose an <orthonormal basis> in each eigenspace of $T_\rho^*T_\rho$ and put $B_p=T_\rho A_p/\lambda_p$. Therefore, when $\rho_p=\rho_q$,
$$
\boxed{\operatorname{tr}(U(p)^*U(q))
=\operatorname{tr}(V(p)^*V(q))
=n_p\delta_{pq}}.
$$
This is (iv), with the unnormalized <Hilbert-Schmidt inner product> and the <Kronecker delta>. No orthogonality of differently shaped matrices is being asserted.
For the unheaded continuation, partition $b\in\mathbb C^m$ into blocks $b_p\in\mathbb C^{n_p}$. Then
$$
UP(x)^*b=\sum_pU(p)\rho_p(x)^*b_p.
$$
The <Schur averaging of rectangular matrices> formula is
$$
\mathbb E_x\rho_p(x)M\rho_q(x)^*
=\begin{cases}(\operatorname{tr}M/n_p)I_{n_p},&\rho_p=\rho_q,\\0,&\rho_p\ne\rho_q,\end{cases}
$$
for any $n_p\times n_q$ matrix $M$. The average is an <intertwiner>; the <Schur lemma> makes it zero for inequivalent <irreducible representations>, and a scalar multiple of the identity for the same representative. The <trace> determines that scalar in the latter case.
Apply this with $M=U(p)^*U(q)$. The <Hilbert-Schmidt inner product> normalization just proved gives
$$
\mathbb E_x\rho_p(x)U(p)^*U(q)\rho_q(x)^*
=\begin{cases}I_{n_p},&p=q,\\0,&p\ne q.\end{cases}
$$
Expanding the squared <Euclidean norm> now yields
$$
\begin{aligned}
\mathbb E_x\|UP(x)^*b\|_2^2
&=\sum_{p,q}b_p^*\mathbb E_x\bigl[\rho_p(x)U(p)^*U(q)\rho_q(x)^*\bigr]b_q\\
&=\sum_p\|b_p\|_2^2
=\boxed{\|b\|_2^2}.
\end{aligned}
$$
Thus the final identity follows from averaging, although $U^*U$ itself need not be the identity. The whole construction is the <spectral inverse theorem for the matrix-valued U2 quantity>.
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