= Solution
Write $V=V_0e^{-A\phi/m_{\rm pl}}$, where the stated model has $V_0=m_{\rm pl}^4$, and take $A>0$. Slow roll gives
$$
H=\sqrt{\frac{8\pi V_0}{3m_{\rm pl}^2}}e^{-A\phi/(2m_{\rm pl})},\qquad\dot\phi=\frac{A\sqrt{V_0}}{\sqrt{24\pi}}e^{-A\phi/(2m_{\rm pl})}.
$$
Thus $d[e^{A\phi/(2m_{\rm pl})}]/dt=A^2\sqrt{V_0}/(\sqrt{96\pi}m_{\rm pl})$. Absorb the integration constant into $b$ to obtain
$$
\boxed{\phi(t)=\frac{2m_{\rm pl}}A\log[C(t+b)],\qquad C^2=\frac{A^4V_0}{96\pi m_{\rm pl}^2}}.
$$
It follows that $H=16\pi/[A^2(t+b)]$, so
$$
\boxed{\frac{a(t)}{a(t_i)}=\left(\frac{t+b}{t_i+b}\right)^p=\exp\left[\frac{8\pi}{Am_{\rm pl}}(\phi-\phi_i)\right],\qquad p=\frac{16\pi}{A^2}}.
$$
The exponential potential has constant $\epsilon_V=A^2/(16\pi)$ and $\eta_V=A^2/(8\pi)$. Inflation requires $p>1$, whereas reliable <slow-roll approximation> requires $A^2/(8\pi)\ll1$. This is the <slow-roll normalization for exponential inflation>.
The stated field normalization is a slow-roll result, not an exact solution of the original full equations. Keeping the acceleration and kinetic terms yields the same $p$ but $C^2=A^4V_0/[2m_{\rm pl}^2(48\pi-A^2)]$, as follows by substituting $\dot\phi=2m_{\rm pl}/[A(t+b)]$ into both full equations. It agrees to leading order for small slope and is the <exponential-potential power-law inflation> solution.
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