Step 1 — Examine the Assertion (A).

The sample space of a single throw of a die is $S = \{1, 2, 3, 4, 5, 6\}$.

Let

• $A$ = the number is prime = $\{2, 3, 5\}$
• $B$ = the number is odd = $\{1, 3, 5\}$

Step 2 — Find the required conditional probability.

$$A \cap B = \{3, 5\} \quad \Longrightarrow \quad P(A \cap B) = \frac{2}{6}$$

$$P(B) = \frac{3}{6}$$

$$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{2/6}{3/6} = \frac{2}{3}$$

This matches the claim, so Assertion (A) is TRUE.

Step 3 — Examine the Reason (R).

The correct definition of conditional probability is

$$P(A|B) = \frac{P(A \cap B)}{P(B)}, \qquad P(B) \ne 0$$

The Reason states $\dfrac{P(A \cup B)}{P(B)}$, using union instead of intersection. This is incorrect.

So Reason (R) is FALSE.

Step 4 — Conclusion.

Assertion is true and Reason is false.

$$\boxed{\text{Option (C)}}$$