= Solution
Apply the stated deterministic equivalent for the squared resolvent with $\Theta_p=\beta^0(\beta^0)^T/r^2$. The bias term tends to $-\ell^2r^2m'(\ell)$. For the variance use
$$
\widehat\Sigma(\widehat\Sigma+\ell I)^{-2}
=(\widehat\Sigma+\ell I)^{-1}-\ell(\widehat\Sigma+\ell I)^{-2}.
$$
Normalized traces and $p/n\to\gamma$ therefore give
$$
R_X(\widehat\beta_\lambda)
\xrightarrow{\mathrm{a.s.}}
-\ell^2r^2m'(\ell)+\sigma^2\gamma\bigl(m(\ell)+\ell m'(\ell)\bigr).
$$
Thus the conclusion printed in the paper also has both signs involving $m'$ reversed. The hypotheses as printed imply the formula above; they cannot imply the requested one because $(\widehat\Sigma+\ell I)^{-2}$ is positive definite while $m'(\ell)\leq0$ for a resolvent limit.
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