Step 1: List the digits available.
The single digit primes are 2, 3, 5, and 7, so there are 4 digits to choose from in total.
Step 2: Fill the 3-digit number one position at a time, using the multiplication principle.
Instead of splitting the problem into "choose the set" and "arrange the set" separately, we can directly count how many digits are available for each of the three positions (hundreds, tens, units) of the number, one after another.
Step 3: Fill the hundreds place.
Any of the 4 primes (2, 3, 5, 7) can go in the hundreds place, since none of them is zero and all are allowed to lead the number. This gives 4 choices for the first digit.
Step 4: Fill the tens place.
The digits must all be distinct, so the tens digit cannot repeat whichever digit was used in the hundreds place. That leaves $4 - 1 = 3$ digits still available for the tens place.
Step 5: Fill the units place.
By the same logic, two digits are already used up in the hundreds and tens places, so only $4 - 2 = 2$ digits remain available for the units place.
Step 6: Multiply the choices for each position.
\[ 4 \times 3 \times 2 = 24 \]
Each step's choice is independent of how we count it, whether by first fixing the set of digits or by filling positions left to right, so this matches the combinations-then-arrangements method exactly.
Step 7: Confirm the wrong options.
Getting 64 (option A) would mean allowing a digit to be reused in more than one position, which is ruled out by the word "distinct" in the question. Getting 12 or 4 (options C and D) means one of the three position-fill steps was skipped or miscounted.
Final Answer:
Filling the hundreds, tens, and units positions one at a time with 4, then 3, then 2 remaining digits gives $4 \times 3 \times 2 = 24$.
\[ \boxed{24} \]