= Solution
The <slow-roll equations> give $H^2\simeq V/(3M_{\mathrm{Pl}}^2)$ and $\dot\Phi\simeq-V'/(3H)$. Substituting in the stated <slow-roll curvature power spectrum> yields
$$
\mathcal P_{\mathcal R}=\frac{H^4}{4\pi^2\dot\Phi^2}
\simeq\frac{9H^6}{4\pi^2V'^2}
=\frac{V^3}{12\pi^2M_{\mathrm{Pl}}^6V'^2}.
$$
With the <potential slow-roll parameter> $\epsilon_V=M_{\mathrm{Pl}}^2V'^2/(2V^2)$,
$$
\boxed{\mathcal P_{\mathcal R}\simeq\frac{V}{24\pi^2M_{\mathrm{Pl}}^4\epsilon_V}
=\frac8{3\epsilon_V}\left(\frac{V^{1/4}}{\sqrt{8\pi}\,M_{\mathrm{Pl}}}\right)^4.}
$$
All quantities are evaluated at horizon exit. The two forms have the same normalization because $(\sqrt{8\pi})^4=64\pi^2$.
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