$P(x)-a=(x-1)(x-3)$ $\cdot(x-5)(x-7)Q(x)$ for some integer polynomial $Q$

Evaluate at $x=2,4,6,8$ to force $2a$ to be divisible by $\operatorname{lcm}(15,9,105)$

Since $1, 3, 5, 7$ are roots of $P(x)-a,$ write $P(x)-a=(x-1)(x-3)$ $\cdot(x-5)(x-7)Q(x)$ with $Q$ having integer coefficients.

Evaluating at $x=2, 4, 6, 8$ (where $P=-a$) gives $$\begin{aligned} -2a &=-15\,Q(2) \\ &=9\,Q(4) \\ &=-15\,Q(6) \\ &=105\,Q(8). \end{aligned}$$

So $15, 9,$ and $105$ all divide $2a,$ hence $\operatorname{lcm}(15,9,105)=315$ divides $2a.$ Since $315$ is odd, $a$ is divisible by $315,$ so $a\ge315.$

To attain the bound, let $$\begin{aligned} F(x)&=(x-1)(x-3) \\ &\quad\cdot(x-5)(x-7), \\ Q(x)&=42+(x-2)(x-6) \\ &\quad\cdot(60-8x), \\ P(x)&=315+F(x)Q(x). \end{aligned}$$ This integer polynomial equals $315$ at $1,3,5,7.$ At $2,4,$ the pairs $(F(x),Q(x))$ are $(-15,42),(9,-70),$ and at $6,8$ they are $(-15,42),(105,-6).$ Thus $F(x)Q(x)=-630$ each time and $P(x)=-315.$ Hence the bound is attainable.

Thus, the correct answer is B.