Set A is not computable.
Proof. A is nontrivial because some programs satisfy the required property for membership in A, and some do not.
A respects equivalence. Suppose that p and q are two equivalent programs.
p ∈ A | ⇒ | φp(n)↓ when n is an even number and φp(n)↑ when n is an odd number | |
⇒ | φq(n)↓ when n is an even number and φp(n)↑ when n is an odd number | (since p ≈ q) | |
⇒ | q ∈ A |
By Rice's theorem, B is uncomputable.