Periodogram Averaging
Since the data records $x_i$ are uncorrelated realizations of the same random process, $\hat P^{per}_{(i)}(e^{j\omega})$ are all uncorrelated random variables with identical means and variances (given by the mean and variance of the Bartlett Method's PSD estimator). So, for any fixed $\omega$, $\mathbf E[\hat P^{per}_{(i)}(e^{j\omega})]$ are equal $\forall i$ and also $\mathrm{Var}[\hat P^{per}_{(i)}(e^{j\omega})]$ are equal $\forall i$. So,
\begin{eqnarray}
\mathbf E[\hat P_{x}(e^{j\omega})] &amp;=&amp; \mathbf E\left[\frac{1}{K} \sum_{i=1}^K\hat P^{per}_{(i)}(e^{j\omega})\right] \\
&amp;=&amp; \frac{1}{K}\sum_{i=1}^K \mathbf E\left[\hat P^{per}_{(i)}(e^{j\omega})\right] \\
&amp;=&amp; \frac{1}{K} K \;\mathbf E\left[\hat P^{per}_{(i)}(e^{j\omega})\right]\\
&amp;=&amp; \mathbf E\left[\hat P^{per}_{(i)}(e^{j\omega})\right]
\end{eqnarray}

and similarly,

\begin{eqnarray}
\mathrm{Var}[\hat P_{x}(e^{j\omega})] &amp;=&amp; \mathrm{Var}\left[\frac{1}{K} \sum_{i=1}^K\hat P^{per}_{(i)}(e^{j\omega})\right] \\
&amp;=&amp; \frac{1}{K^2}\sum_{i=1}^K \mathrm{Var}\left[\hat P^{per}_{(i)}(e^{j\omega})\right] \\
&amp;=&amp; \frac{1}{K^2} K \;\mathrm{Var}\left[\hat P^{per}_{(i)}(e^{j\omega})\right]\\
&amp;=&amp; \frac{1}{K}\mathrm{Var}\left[\hat P^{per}_{(i)}(e^{j\omega})\right].
\end{eqnarray}