If $$A$$ and $$B$$ are two events such that $$P(A \cup B) = P(A \cap B)$$, then the incorrect statement amongst the following statements is:

We are given that for two events $$A$$ and $$B$$, $$P(A \cup B) = P(A \cap B)$$. Let $$P(A \cap B) = x$$. Then, $$P(A \cup B) = x$$.

Recall the formula for the probability of the union of two events: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$. Substituting the given values, we get:

$$x = P(A) + P(B) - x$$

Adding $$x$$ to both sides:

$$x + x = P(A) + P(B)$$

$$2x = P(A) + P(B)$$

So, $$P(A) + P(B) = 2x$$, which means $$P(A) + P(B) = 2P(A \cap B)$$. Let this be equation (1).

Now, we examine each option to find which one is incorrect.

Option A: $$P(A) + P(B) = 1$$

From equation (1), $$P(A) + P(B) = 2P(A \cap B)$$. This equals 1 only if $$2P(A \cap B) = 1$$, i.e., $$P(A \cap B) = \frac{1}{2}$$. However, this is not necessarily true. For example, if $$A$$ and $$B$$ are the entire sample space, then $$P(A) = 1$$, $$P(B) = 1$$, $$P(A \cap B) = 1$$, and $$P(A \cup B) = 1$$, so $$P(A \cup B) = P(A \cap B)$$ holds, but $$P(A) + P(B) = 2 \neq 1$$. Thus, this statement is not always true.

Option B: $$P(A \cap B') = 0$$

Note that $$A \cap B'$$ is the set of elements in $$A$$ but not in $$B$$. We can express $$P(A)$$ as:

$$P(A) = P(A \cap B) + P(A \cap B')$$

Similarly, for $$B$$:

$$P(B) = P(A \cap B) + P(A' \cap B)$$

Substituting into equation (1):

$$P(A) + P(B) = [P(A \cap B) + P(A \cap B')] + [P(A \cap B) + P(A' \cap B)] = 2P(A \cap B) + P(A \cap B') + P(A' \cap B)$$

But from equation (1), $$P(A) + P(B) = 2P(A \cap B)$$, so:

$$2P(A \cap B) + P(A \cap B') + P(A' \cap B) = 2P(A \cap B)$$

Subtracting $$2P(A \cap B)$$ from both sides:

$$P(A \cap B') + P(A' \cap B) = 0$$

Since probabilities are non-negative, each term must be zero. Therefore, $$P(A \cap B') = 0$$ and $$P(A' \cap B) = 0$$. Hence, this statement is true.

Option C: $$A$$ and $$B$$ are equally likely

This means $$P(A) = P(B)$$. From option B, we have $$P(A \cap B') = 0$$ and $$P(A' \cap B) = 0$$. This implies that the event $$A - B$$ (elements in $$A$$ but not in $$B$$) has probability zero, and similarly, $$B - A$$ has probability zero. Therefore, $$A$$ and $$B$$ are equal except possibly on a set of probability zero, so $$P(A) = P(B)$$. Thus, this statement is true.

Option D: $$P(A' \cap B) = 0$$

As derived in option B, $$P(A' \cap B) = 0$$. Hence, this statement is true.

Only option A is not always true, as it depends on the specific value of $$P(A \cap B)$$. Therefore, the incorrect statement is option A.

Hence, the correct answer is Option A.