Solution:

Because $1,3,5,7$ are roots of $P(x)-a$, write $P(x)-a=(x-1)(x-3)(x-5)(x-7)Q(x)$, where $Q(x)$ has integer coefficients.

Substituting $x=2,4,6,8$ gives $-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8)$. Hence $a$ must be a multiple of $\operatorname{lcm}(15,9,105)=315$.

This lower bound is attainable: take $Q(x)=42+(x-2)(x-6)(60-8x)$ and define $P(x)=315+(x-1)(x-3)(x-5)(x-7)Q(x)$. This polynomial has integer coefficients and satisfies the required values.

Thus, B is the correct answer.