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 as $$ y_{it}=\frac{\alpha_i}{1-\gamma} +\sum_{k=0}^{\infty} \varepsilon_{it-k} \cdot \gamma^k $$ Using the formula 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 $$Cov(y_{it},y_{it-1})=Cov(\gamma \cdot y_{it-1}+\alpha_i + \varepsilon_{it},y_{it-1})=\gamma \cdot Cov(y_{it-1},y_{it-1})+Cov(\alpha_i,y_{it-1})+0=\gamma \cdot V(y_{it-1})+Cov(\alpha_i,\frac{\alpha_i}{1-\gamma} +\sum_{k=0}^{\infty} \varepsilon_{it-k} \cdot \gamma^k) = \frac{\sigma^2_{\alpha}}{1-\gamma} $$
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.