To determine which of the given languages is NOT regular, we need to analyze each option considering that $ L $ is a regular language over the alphabet $ \Sigma = \{ a, b \} $.

**Option A: Suffix $ (L) = \{ y \in \Sigma^* \mid \exists x \in \Sigma^*, xy \in L \} $**

Explanation:

Suffixes of a Regular Language:

The set of all suffixes of strings in a regular language $ L $ is regular.

Regular languages are closed under quotient operations, including taking suffixes.

Construction:

Given a DFA for $ L $, we can construct an NFA that accepts all suffixes by:

Including epsilon transitions from the start state to all other states.

This allows the NFA to "jump" into any point in the DFA and accept suffixes.

Conclusion:

Suffix $ (L) $ is regular.

Option B: $ \{ ww^R \mid w \in L \} $

Explanation:

Understanding $ ww^R $:

$ w^R $ is the reverse of string $ w $.

The language consists of strings where a word from $ L $ is concatenated with its reverse.

Regularity Analysis:

Even if $ L $ is regular, $ \{ ww^R \mid w \in L \} $ is not necessarily regular.

Known Non-Regular Language:

When $ L = \Sigma^* $, $ \{ ww^R \mid w \in L \} = \{ ww^R \mid w \in \Sigma^* \} $ is a classic example of a non-regular language.

Reason:

Regular languages cannot enforce the condition that a string is followed by its reverse due to their limited memory (cannot count or compare arbitrary lengths).

Closure Properties:

Regular languages are not closed under certain operations, including some forms of concatenation that require matching patterns (like palindromes).

Conclusion:

$ \{ ww^R \mid w \in L \} $ is not necessarily regular and, in general, is non-regular.

**Option C: Prefix $ (L) = \{ x \in \Sigma^* \mid \exists y \in \Sigma^*, xy \in L \} $**

Explanation:

Prefixes of a Regular Language:

The set of all prefixes of strings in a regular language $ L $ is regular.

Regular languages are closed under taking prefixes.

Construction:

Given a DFA for $ L $, we can construct an NFA that accepts all prefixes by:

Modifying the DFA so that any state that is reachable becomes an accepting state.

This accepts any string that leads to any state in the DFA.

Conclusion:

Prefix $ (L) $ is regular.

Option D: $ L \cdot L^R = \{ xy \mid x \in L, y^R \in L \} $

Explanation:

Reversal of Regular Language:

The reversal $ L^R $ of a regular language $ L $ is regular.

Regular languages are closed under reversal.

Concatenation:

The concatenation of two regular languages is regular.

Since both $ L $ and $ L^R $ are regular, their concatenation $ L \cdot L^R $ is regular.

Conclusion:

$ L \cdot L^R $ is regular.

Final Answer:

Among the given options, Option B represents a language that is not regular.

Answer: Option B