= Solution
Use $\nu_T=u_*\kappa y$, $\epsilon=u_*^3/(\kappa y)$, $k=u_*^2/\sqrt{C_\mu}$, and $\mathcal P=\epsilon$ in the dissipation equation. Its diffusion term is
$$
\frac{d}{dy}\left(\frac{\nu_T}{\sigma_\epsilon}\frac{d\epsilon}{dy}\right)=\frac{d}{dy}\left(-\frac{u_*^4}{\sigma_\epsilon y}\right)=\frac{u_*^4}{\sigma_\epsilon y^2}.
$$
The net modeled source is $(C_{\epsilon1}-C_{\epsilon2})\epsilon^2/k=-(C_{\epsilon2}-C_{\epsilon1})\sqrt{C_\mu}\,u_*^4/(\kappa^2y^2)$. Dividing the equation by $u_*^4/y^2$ and solving gives
$$
\boxed{\kappa^2=\sigma_\epsilon\sqrt{C_\mu}(C_{\epsilon2}-C_{\epsilon1}),\qquad\kappa=C_\mu^{1/4}\sqrt{\sigma_\epsilon(C_{\epsilon2}-C_{\epsilon1})}.}
$$
The physical <Von Kármán constant> is the positive root, requiring $C_{\epsilon2}>C_{\epsilon1}$ with positive $C_\mu,\sigma_\epsilon$. Neglecting dissipation diffusion would incorrectly require the two dissipation-equation constants to be equal. The parameter $\sigma_k$ does not enter because the constant $k$ profile has no energy-diffusion contribution.
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