Variance of white noise and the STFT
$E\{|D_k|^2\}$ is the expected value of the squared magnitude of the noise in the $k^{th}$ frequency bin. Since they assume that the noise is zero mean, this equals the noise variance. Remember that the variance of a random variable $X$ is defined as $\sigma_X^2=E\{|X-\mu_X|^2\}$, where $\mu_X=E\{X\}$ is the mean of $X$. So if $\mu_X=0$ (as assumed here for the noise) then $\sigma^2_X=E\{|X|^2\}$. Note that $E\{|D_k|^2\}$ is also the noise power in the $k^{th}$ bin, so it can't be zero unless there is no noise at all in that bin.