Calculating Power of Sinusoidal Term and a Convolution expression The following is an excerpt from the book Statistical Digital Signal Processing and Modelling (Monson Hayes)


  ...consider a random process consisting of a random phase sinusoid in
  white noise $$ x(n) = A\sin(n\omega_{0} + \phi) + v(n)$$ where $\phi $
  is a random variable that is uniformly distributed over the interval
  $[-\pi, \pi]$, and $v(n)$ is white noise with a variance of
  $\sigma_{v}^{2}$. The power spectrum of $x(n)$ is 
  
  $$ P_{x}(e^{j\omega}) =  \sigma_{v}^{2} + \frac{1}{2}\pi
 A^{2}[u_{0}(\omega - \omega_{0}) + u_{0}(\omega + \omega_{0})]$$ 
  
  Therefore it follows from Eq. (8.23) that the expected value of the
  periodogram is  $$E\{{P_{per}(e^{j\omega})}\} =
 \frac{1}{2\pi}P_{x}(e^{j\omega})*W_{B}(e^{j\omega})$$ $$=
 \sigma_{v}^{2} + \frac{1}{4}A^{2}[W_{B}(e^{j(\omega - \omega_{0})}) +
 W_{B}(e^{j(\omega + \omega_{0})})]$$



I am unable to figure out how he obtained the expression for power of the sinusoidal term. Shouldn't it be just $\pi^{2}A^{2}$ multiplied by the bracket term? 
Considering that $W_{B}$ represents fourier transform of Bartlett window, how is the convolution expression being evaluated here?
The power spectrum $P(e^{j\omega})$ is the Fourier transform of the autocorrelation function of $x(n)$:

$$R_x(k)=E\{x(n)x(n+k)\}\tag{1}$$

If you evaluate $(1)$ you should get

$$R_x(k)=\sigma_v^2\delta(k)+\frac{A^2}{2}\cos(k\omega_0)\tag{2}$$

Taking the Fourier transform of $(2)$ leads to the given expression for $P(e^{j\omega})$.

The convolution is obtained by noting that

$$u_0(\omega-\omega_0)\star W_B(e^{j\omega})=W_B(e^{j(\omega-\omega_0)})\tag{3}$$

and

$$\frac{1}{2\pi}\sigma_v^2\star W_B(e^{j\omega})=\sigma_v^2\frac{1}{2\pi}\int_{-\pi}^{\pi}W_B(e^{j\omega})d\omega=\sigma_v^2\tag{4}$$