Real coefficients pair nonreal roots as conjugates, so $P(z)$ factors into linear $(z - c)$ and quadratic $z^2 - 2az + (a^2 + b^2)$ pieces, each with constant term dividing $50$

For each divisor $d$ of $50,$ count the basic factors of magnitude $d$ (solutions of $a^2 + b^2 = d,$ plus $z \pm d$), then multiply choices so the constant terms multiply to $50$

Since the coefficients are real, nonreal roots occur in conjugate pairs, so $P(z)$ factors into distinct linear factors $(z - c)$ with $c \in \mathbb{Z}$ and quadratics $(z - (a+ib))(z - (a-ib))$ $= z^2 - 2az + (a^2 + b^2).$ Each factor’s constant term divides $50.$ For $d=1,2,5,10,25,50,$ the numbers of conjugate pairs with $a^2+b^2=d$ and $b\ne0$ are $1,2,4,4,5,6,$ respectively. Adding the two linear choices $z-d,z+d$ gives $|B_1|=3,$ $|B_2|=4,$ $|B_5|=6,$ $|B_{10}|=6,$ $|B_{25}|=7,$ and $|B_{50}|=8.$ The factor-magnitude partitions of $50$ using values greater than $1$ are $50,25\cdot2,10\cdot5,$ and $5\cdot5\cdot2.$ Distinct roots require choosing two different $B_5$ factors in the last case. Finally, account for the free presence of $z+1$ and $z^2+1$ (with $z-1$ forced by the sign of the remaining product), gives $$\begin{aligned} &2^2\left(8 + 7\cdot 4 + 6\cdot 6 + 4\binom{6}{2}\right) \\ &\quad = 4(8 + 28 + 36 + 60) = 528. \end{aligned}$$ Thus, the correct answer is B.