Step 1: Identify the pool of digits.
Single digit primes are 2, 3, 5, 7, so the pool has $n = 4$ digits, and we must arrange $r = 3$ of them, in order, with no repetition, to build each 3-digit number.
Step 2: Use the permutation formula directly.
The number of ways to arrange $r$ items chosen from $n$ distinct items, where order matters and repetition is not allowed, is given by
\[ P(n, r) = \frac{n!}{(n-r)!} \]
Substitute $n = 4$ and $r = 3$:
$P(4, 3) = \dfrac{4!}{(4-3)!} = \dfrac{4!}{1!} = \dfrac{24}{1} = 24$
Step 3: Sanity check with a small direct count.
Pick any 3 of the 4 primes, say 2, 3, 5, leaving out 7. These three digits alone can be arranged in $3! = 6$ different orders (235, 253, 325, 352, 523, 532). There are $\binom{4}{3} = 4$ different ways to choose which 3 of the 4 primes to leave out, and each such choice gives 6 orderings, so the total is $4 \times 6 = 24$, matching the permutation formula.
Step 4: Match to the given options.
The computed count of 24 matches option (B) exactly; 64 assumes repeated digits are allowed, and 12 and 4 both undercount the true number of orderings.
Final Answer:
There are 24 such 3-digit numbers, option (B).