Solution 1 (by Titu Andreescu). Using Vieta's identities we have:
$$x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4 - x_1x_2x_3x_4 \ge 5,$$
and so
$$x_1(x_2 + x_3 + x_4 - x_2x_3x_4) + 1(x_2x_3 + x_2x_4 + x_3x_4 - 1) \ge 4.$$
It follows that
$$4^2 \le [x_1(x_2 + x_3 + x_4 - x_2x_3x_4) + 1(x_2x_3 + x_2x_4 + x_3x_4 - 1)]^2,$$
so by the Cauchy-Schwarz Inequality,
$$\begin{aligned} 4^2 &\le (x_1^2 + 1)[(x_2 + x_3 + x_4 - x_2x_3x_4)^2 + (x_2x_3 + x_2x_4 + x_3x_4 - 1)^2] \\ &= (x_1^2 + 1)(x_2^2 + 1)(x_3^2 + 1)(x_4^2 + 1). \end{aligned}$$
Setting $x_1 = \dots = x_4 = 1$ gives $b - d = 5$ and makes the product attain its minimum possible value of $16$.

Solution 2 (by Titu Andreescu). An alternative, shorter argument runs as follows: we have
$$(x_1^2 + 1)(x_2^2 + 1)(x_3^2 + 1)(x_4^2 + 1) = P(i)P(-i) = ((1 - b + d) + i(c - a))(1 - b + d - i(c - a)) = (b - d - 1)^2 + (c - a)^2 \ge 16,$$
with equality if and only if $b - d = 5$ and $a = c$, both attained if $x_1 = \dots = x_4 = 1$.