Step 1: Assign shorthand
Let us denote the puppies as A (Ashlen), B (Blakely), C (Custard), D (Daffy), E (Earl), F (Fala), G (Gabino).
Step 2: Apply given conditions
- If G is in Pen 1, then D is not in Pen 2 \(\Rightarrow\) G and D must be together in Pen 1.
- If D is not in Pen 2, then G must be in Pen 1 \(\Rightarrow\) confirms both G and D in Pen 1.
- If A is in Pen 2, then B cannot be in Pen 2 \(\Rightarrow\) A and B cannot be in the same pen.
- If B is in Pen 1, then A cannot be in Pen 1 \(\Rightarrow\) again, A and B must be in opposite pens.
Step 3: Construct possibilities
Pen 1 must contain G and D together. The third member of Pen 1 could be either A or B (depending on constraints).
Checking all valid distributions, only in Case II (from the explanation table), Pen 2 is \(\{B, C, E, F\}\) \(\Rightarrow\) Blakely, Custard, Earl, Fala.
Step 4: Match with options
This matches exactly with Option D.
\[
\boxed{\text{Blakely, Custard, Earl, Fala}}
\]