Let $p \equiv 2 \pmod 3$ be a prime, $k$ a positive integer and $P(x) = 3x^{\frac{2p-1}{3}}+3x^{\frac{p+1}{3}}+x+1$. For any integer $n$, let $R(n)$ denote the remainder when $n$ is divided by $p$ and let $S = \{0,1,\cdots,p-1\}$. At each step, you can either (a) replaced every element $i$ of $S$ with $R(P(i))$ or (b) replaced every element $i$ of $S$ with $R(i^k)$. Determine all $k$ such that there exists a finite sequence of steps that reduces $S$ to $\{0\}$.

[i]Proposed by fattypiggy123[/i]
1. **Observation of Polynomial $P(x)$:**
   We start by analyzing the polynomial \( P(x) = 3x^{\frac{2p-1}{3}} + 3x^{\frac{p+1}{3}} + x + 1 \). Notice that for a prime \( p \equiv 2 \pmod{3} \), we can rewrite \( P(x) \) in a simpler form modulo \( p \).

2. **Simplification of $P(x)$:**
   Observe that:
   \[
   P(x) = 3x^{\frac{2p-1}{3}} + 3x^{\frac{p+1}{3}} + x + 1
   \]
   can be rewritten as:
   \[
   P(x) \equiv (x^{\frac{1}{3}} + 1)^3 \pmod{p}
   \]
   This is because \( p \equiv 2 \pmod{3} \) implies that \( 3 \) is invertible modulo \( p \), and the exponents simplify accordingly.

3. **Properties of the Map $x \rightarrow x^3$:**
   The map \( x \rightarrow x^3 \) is bijective modulo \( p \) because \( p \) is a prime. This means every element in \( \mathbb{Z}_p \) has a unique cube root modulo \( p \).

4. **Defining the Set $T$:**
   Define the set \( T = \{ x^{\frac{1}{3}} \mid x \in S \} \). The first operation \( x \rightarrow P(x) \) translates to adding 1 to each element of \( T \), and the second operation \( x \rightarrow x^k \) raises each element of \( T \) to the power of \( k \).

5. **Behavior of Operations on $T$:**
   - The first operation \( x \rightarrow P(x) \) effectively becomes \( x \rightarrow x + 1 \) in the set \( T \).
   - The second operation \( x \rightarrow x^k \) raises each element of \( T \) to the power of \( k \).

6. **Conditions for Reducing $S$ to $\{0\}$:**
   To reduce \( S \) to \( \{0\} \), we need to ensure that the set \( T \) can be reduced to a single element. This is possible if and only if \( \gcd(k, p-1) > 1 \). 

   - If \( \gcd(k, p-1) = 1 \), the map \( x \rightarrow x^k \) is bijective, and \( T \) never changes in size.
   - If \( \gcd(k, p-1) > 1 \), there exists a positive integer \( a \) such that \( p \mid a^k - 1 \) and \( p \nmid a - 1 \). This allows us to reduce the size of \( T \) by applying the operations repeatedly.

7. **Reduction Process:**
   Suppose \( |T| \ge 2 \) at some point. Let \( d \) be the difference between two distinct elements of \( T \). Since \( p \nmid a - 1 \), there exists an \( x \) such that \( ax \equiv x + d \). By applying the first operation sufficiently many times, we can ensure \( x, ax \in T \). Applying the second operation \( x \rightarrow x^k \) will then reduce the size of \( T \).

8. **Conclusion:**
   Repeating the above process will eventually reduce \( T \) to a single element, thus reducing \( S \) to \( \{0\} \).

The final answer is \( \boxed{ \gcd(k, p-1) > 1 } \).