N=8
For k=5, $X(5) = x*(8-5) \\
= x*(3) = 0$
For k=6, $X(6) = x*(8-6) \\
= x*(2) = (-4-4j)$
For k=7, $X(7) = x*(8-7) \\
= x*(1) = 0$
$\therefore, X(5) = 0, X(6) = -4-4j, X(7) = 0$
Result: X(K)={20, 0, -4+4j, 0, -4, 0, -4-4j, 0}
(i) Let $x_1(n)=(-1)n$
$\therefore$
$\therefore, x_1$ (n)={1, -1, 1, -1, 1, -1, 1, -1}
$\therefore, X_1$(k)={0, 0, 0, 0, 8, 0, 0, 0}
By Convolution Property of DFT, Convolution of two signals in time domain is equivalent to multiplication in frequency domain.
$\therefore, P(K)= X_1(k). X(k) \ \ \ \ as, p(n) = x_1(n) * x(n)$
$\therefore$, By Convolution Theorem,
P(K) = {0, 0, 0, 0, 8, 0, 0, 0}. {20, 0, -4+4j, 0, -4, 0, -4-4j, 0}
Result: P(K)={0, 0, 0, 0, -32, 0, 0, 0}