= Solution
\b[The claimed equality does not hold for the printed joint distribution.] The original PDF specifies a centered <Maxwell-Boltzmann distribution> of $\mathbf v=\dot{\mathbf R}$ independent of $\mathbf R$, with
$$
\langle v_i\rangle=0,\qquad\langle v_iv_j\rangle=\frac{2k_BT}{m}\delta_{ij}.
$$
Since the double average includes number density $n$, evaluating the <Gaussian integrals> gives
$$
\boxed{\left\langle\!\left\langle\frac\zeta2|\mathbf v-K\mathbf R|^2\right\rangle\!\right\rangle
=\frac{3n\zeta k_BT}{m}+\frac{n\zeta}{2}\operatorname{tr}(KCK^T).}
$$
The cross term vanishes because the conditional mean of $\mathbf v$ is zero. In particular, set $K=0$ and take the equilibrium configuration <probability density> $\psi_{\rm eq}\propto e^{-HR^2/(2k_BT)}$. The left side is $3n\zeta k_BT/m>0$, while $nHC:K=0$. This is a counterexample even at thermal equilibrium, so no rearrangement of the earlier <conformation tensor> equation can prove the printed statement.
The discrepancy is <thermal bead drag and heat-bath power>: gross viscous loss of microscopic thermal motion is replenished by the thermal bath. It is not identical to mechanical polymer stress power. Moreover, $\dot{\mathbf R}$ in the earlier configuration continuity equation is a <probability current velocity>, including a distribution-dependent osmotic drift; it cannot simultaneously be identified with an independently sampled centered Maxwellian velocity.
A consistent overdamped result is the <free-energy balance of a Hookean dumbbell>. Define $\mathbf q=H\mathbf R+k_BT\nabla_{\mathbf R}\log\psi$ and
$$
\mathcal F=nk_BT\int\psi\log\frac\psi{\psi_{\rm eq}}\,d\mathbf R,\qquad
\mathbf V=K\mathbf R-\frac2\zeta\mathbf q.
$$
Differentiate the <relative entropy> and use the configuration <Fokker-Planck probability current> $\psi\mathbf V$. After <integration by parts>,
$$
\dot{\mathcal F}=n\int\psi\,\mathbf q\cdot\mathbf V\,d\mathbf R
=nHC:K-\frac{2n}{\zeta}\int\psi|\mathbf q|^2\,d\mathbf R.
$$
The affine osmotic term is $nk_BT\int\nabla\psi\cdot K\mathbf R\,d\mathbf R=-nk_BT\operatorname{tr}K=0$. Hence
$$
\boxed{nHC:K=\dot{\mathcal F}+\mathcal D,\qquad\mathcal D=\frac{2n}{\zeta}\int\psi|\mathbf q|^2\,d\mathbf R\ge0.}
$$
The polymer <stress tensor> $\sigma_p=n(HC-k_BTI)$ has $\sigma_p:K=nHC:K$ in an <incompressible flow>. It stores or releases <free energy> as well as dissipating it, so the total stress-work term in part <2/a> need not itself be a purely dissipative quantity for this <viscoelastic> material. Stress power equals $\mathcal D$ only for stationary <free energy>, and $\mathcal D$ uses the probability-current slip. Neither condition turns the printed independent-Maxwellian average into that expression.
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