a) Since
$ X(e^{j\mu},e^{j\nu}) = \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} x(m,n)e^{-j(m\mu+n\nu)} $
and
$ p_0(e^{jw}) = \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} x(m,n)e^{-jnw} $，
we have:
P0(ej'w) = X(ejμ,ejw) | μ = 0
b) Similarly to a), we have:
$ p_1(e^{jw}) = X(e^{jw}, e^{j\nu})|\nu=0 $
