The problem involves finding the number of subsets of the set $P$ with exactly 2 elements.

Step-by-step solution:

1. Understand the Set $P$: The set $P$ is defined as: $P = \{x \mid x - 3 < 3, x \in \text{Natural Numbers}\}.$ Simplify the inequality $x - 3 < 3$: $x < 6.$ Since $x$ is a natural number ($x \geq 1$), the possible values of $x$ are: $P = \{1, 2, 3, 4, 5\}.$

2. Subsets of $P$ with 2 Elements: To form subsets of $P$ with exactly 2 elements, we calculate the number of combinations of 2 elements from 5 elements: $\binom{n}{r} = \frac{n!}{r!(n - r)!}.$ Here, $n = 5$ and $r = 2$. Substituting these values: $\binom{5}{2} = \frac{5!}{2!(5 - 2)!} = \frac{5 \times 4}{2 \times 1} = 10.$

3. Final Answer: The number of subsets of $P$ with 2 elements is $\mathbf{10}$.

Thus, the correct answer is: $\boxed{\text{C. 10}}$

Do you want further clarification or additional examples?

Expanding the Topic:

1. What is the general formula for calculating subsets with any number of elements?
2. Can you explain how to list all subsets with 2 elements explicitly for the given set $P$?
3. What would the answer be if $P$ contained 6 elements instead of 5?
4. How does the concept of subsets differ when considering different types of sets (e.g., infinite sets)?
5. Can you determine the total number of subsets of $P$?

Tip:

When working with subsets, remember that $\binom{n}{r}$ calculates combinations, which only considers unique selections, not the order of elements.