Just wondering if the following is correct : $$\mathbb P\left(c_1\leqslant\chi^2_{n-p}\leqslant c_2\right)=1-\alpha\\\iff \left\{c_1=\chi^2_{n-p}(1-\alpha/2)\right\}\land\left\{c_2=\chi^2_{n-p}(\alpha/2)\right\}$$
$\text{Edit}$ : A bilateral confidence interval for $\:\sigma^2\:$ at level $\:(1-\alpha)\in\:]0,1[\:$ is $$\left[\frac{\widehat\sigma^2(n-p)}{c_2};\frac{\widehat\sigma^2(n-p)}{c_1}\right].$$
The statement $$\mathbb P\left(c_1\leqslant\chi^2_{n-p}\leqslant c_2\right)=1-\alpha\iff \left\{c_1=\chi^2_{n-p}(1-\alpha/2)\right\}\land\left\{c_2=\chi^2_{n-p}(\alpha/2)\right\}$$ does not make any sense as $\chi^2_{n-p}$ is a continuous random variable, and $c_1=\chi^2_{n-p}(1-\alpha/2)$ occurs with zero probability.
To verify that, consider $c_1 = 0$, $c_2 = \infty$, and $\alpha=0$.