(a) The answer is all $n$ relatively prime to $p-1$. Note that the $n+1$ integers can be reverse constructed from the $n+1$ powers $p^{k_1}, \cdots, p^{k_{n+1}}$ iff $n | \sum p^{k_i}$ If $d>1$ divides both $n$ and $p-1$, then $\sum p^{k_i}\equiv n+1\not\equiv n\equiv 0\pmod{d}$, so $\sum p^{k_i}$ is not a multiple of $n$. If $n$ is relatively prime to $p-1$, let $x\equiv p(p-1)^{-1}\pmod{n}$ such that $0\le x\le n-1$, and take $x$ copies of $1$ and $n+1-x$ copies of $p$. Then $\sum p^{k_i} = x + (n+1-x)p \equiv p - (p-1)x\equiv 0\pmod{n}$ as desired. (b) The answer is all powers of $p$. The condition is strictly stronger than part (a), but with the additional restriction that $\sum p^{k_i} > n\text{max}(p^{k_i})$. WLOG let $k_1\ge k_2\ge \cdots\ge k_{n+1}$. We claim that $k_2$ through $k_n$ are equal to $k_1$. Indeed, assume otherwise, and we have $\sum p^{k_i}\le (n-1)p^{k_1} + 2p^{k_1-1} \le np^{k_1}$, a contradiction. Now, we have $n$ copies of $p^{k_1}$ and some $p^{k_{n+1}}$. The size condition is automatically satisfied, so we just have to worry about mod $n$. However, this simply reduces to $n | np^{k_1} + p^{k_{n+1}}\implies n | p^{k_{n+1}}$, which can happen if and only if $n$ is a power of $p$.