This is a little hand-wavy, but since no one else bit, I will take a stab.
 
You know that
$$
y_{it}=\alpha_i + \varepsilon_{it}+\gamma \cdot y_{it-1}
 =\alpha_i + \varepsilon_{it}+\gamma \cdot (\alpha_i + \varepsilon_{it-1}+\gamma \cdot y_{it-2})
$$
If you keep doing this while assuming that your time series has been running for ever (like time series people always seem to do), you should get something like
$$
y_{it}=\alpha_i \cdot (1+\gamma + \gamma^2 +...) + \sum_{k=0}^{\infty} \varepsilon_{it-k} \cdot \gamma^k
$$
This can be simplified using [the geometric series formula][1] as
$$
y_{it}=\frac{\alpha_i}{1-\gamma} +\sum_{k=0}^{\infty} \varepsilon_{it-k} \cdot \gamma^k
$$
Using [the formula][2] for the variance of sum of weighted random variables (random effects and independence over $t$ assumptions means the covariances are all zero), the variance is then
$$V(y_{it})=\frac{\sigma_{\alpha}^2}{(1-\gamma)^2} +\sigma^2_{\varepsilon} \cdot \sum_{k=0}^{\infty} \gamma^{2k} = \frac{\sigma_{\alpha}^2}{(1-\gamma)^2} + \frac{\sigma_{\varepsilon}^2}{1-\gamma^2}.$$

Getting back to the original question, by definition, the correlation is
$$\rho(y_{it},y_{it-1})=\frac{Cov(y_{it},y_{it-1})}{\sqrt{V(y_{it})\cdot V(y_{it-1})}}$$ 

Stationarity gets you that $V(y_{it})=V(y_{it-1})$, so the denominator simplifies to $V(y_{it}).$ We will use this again below.

The numerator is
\begin{align}
Cov(y_{it},y_{it-1}) & =Cov(\gamma \cdot y_{it-1}+\alpha_i + \varepsilon_{it},y_{it-1}) \\ & =\gamma \cdot Cov(y_{it-1},y_{it-1})+Cov(\alpha_i,y_{it-1})+0 \\ & =\gamma \cdot V(y_{it-1})+Cov \left(\alpha_i,\frac{\alpha_i}{1-\gamma} +\sum_{k=0}^{\infty} \varepsilon_{it-k} \cdot \gamma^k \right) \\ & =\gamma \cdot V(y_{it}) + \frac{\sigma^2_{\alpha}}{1-\gamma}
\end{align}

Putting the top and bottom together, we get
$$\rho(y_{it},y_{it-1})=\frac{\gamma \cdot V(y_{it})+\frac{\sigma^2_{\alpha}}{1-\gamma}}{V(y_{it})}=\gamma+\frac{\frac{\sigma^2_{\alpha}}{1-\gamma}}{ \frac{\sigma_{\alpha}^2}{(1-\gamma)^2} + \frac{\sigma_{\varepsilon}^2}{1-\gamma^2}}$$ 

If you multiply both the top and bottom of that fraction by
$$\frac{(1-\gamma)^2}{\sigma^2_{\alpha}},$$

things will cancel since $1-\gamma^2=(1-\gamma)\cdot(1+\gamma)$, and you will get C&T's expression.


  [1]: https://en.wikipedia.org/wiki/Power_series
  [2]: https://en.wikipedia.org/wiki/Variance#Basic_properties