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:
$ p_0(e^{jw}) = X(e^{j\mu},e^{jw}) |_{\mu=0} $
b) Similarly to a), we have:
$ p_1(e^{jw}) = X(e^{jw},e^{j\nu}) |_{\nu=0} $
c)
$ \sum_{n=-\infty}^{\infty} p_0(n) = \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} x(m,n) = X(e^{j\mu}, e^{j\nu}) |_{\mu=0, \nu=0} $
which is the DC point of the image.
d) No, it can't provide sufficient information.
From the expression in a) and b), we see that $ p_0(e^{jw}) $ and $ p_1(e^{jw}) $ are only slices of the DSFT. It lost the information when $ \mu $ and $ \nu $ are not zero.
A simple example would be:
Let
$ x(m,n) = \left[ {\begin{array}{*{20}{c}} 1 ~ 2 \\ 3 ~ 4\\ \end{array}} \right] $,
so
$ p_0(n) =[4~6], p_1(m) = [3 ~7]^T $.
With the above the information of the projection, the original form of the 2D signal cannot be determined. For example,
$ x(m,n) = \left[ {\begin{array}{*{20}{c}} 2 ~ 1 \\ 2 ~ 5\\ \end{array}} \right] $ gives the same projection.
Related problem: Let $ g(x,y) = sinc(x/2, y/2) $, and let $ s(m,n) = g(mT, nT) $ where T = 1.
a) Calculate $ G(\mu, \nu) $ the CSFT of $ g(x,y) $.
b) Calculate $ S(e^{j\mu}, e^{j\nu}) $ the DSFT of $ s(m,n) $.
