First, we consider the cases where $ z_0 = 1 $ and $ z_0 = -1 $. Note that \[P(1) = 4 + a + b + c + d \ge 4,\]so $ z = 1 $ cannot be a root of $ P(z) $. If $ z = -1 $ is a root of $ P(z), $ then \[P(-1) = 4 - a + b - c + d = (4 - a) + (b - c) + d = 0.\]But $ 4 - a \ge 0, $ $ b - c \ge 0, $ and $ d \ge 0, $ so we must have $ a = 4, $ $ b = c, $ and $ d = 0 $. Conversely, if $ a = 4, $ $ b = c, $ and $ d = 0, $ then \[P(-1) = 4 - a + b - c + d = (4 - a) + (b - c) + d = 0,\]so $ z = -1 $ is a root. In this case, \[P(1) = 4 + a + b + c + d = 4 + 4 + b + b = 8 + 2b.\]The sum of all possible values of $ P(1) $ are then \[\sum_{b = 0}^4 (8 + 2b) = 60.\]Having exhausted the cases where $ z_0 = 1 $ or $ z_0 = -1, $ we can then assume that $ z_0 $ is not real. Let $ z_0 = x_0 + iy_0, $ where $ x_0 $ and $ y_0 $ are real numbers, $ y_0 \neq 0 $. Since $ |z_0| = 1, $ $ x_0^2 + y_0^2 = 1 $. And since the coefficients of $ P(z) $ are real, $ x_0 - iy_0 $ must also be a root, so \[(z - x_0 - iy_0)(z - x_0 + iy_0) = z^2 - 2x_0z + x_0^2 + y_0^2 = z^2 - 2x_0 z + 1\]must be a factor of $ P(z) $. Then \[P(z) = (z^2 - 2x_0 z + 1)(4z^2 + pz + d)\]for some real number $ p $. Expanding, we get \[P(z) = 4z^4 + (p - 8x_0) z^3 + (d - 2px_0 + 4) z^2 + (p - 8x_0) z + d.\]Comparing coefficients, we get \begin{align*} p - 8x_0 &= a, \\ d - 2px_0 + 4 &= b, \\ p - 2dx_0 &= c.\end{align*}Subtracting the first and third equations, we get $ 2dx_0 - 8x_0 = a - c, $ so \[2(d - 4) x_0 = a - c.\quad (*)\]If $ d = 4, $ then $ a = c $. In fact, the chain $ d \le c \le b \le a \le 4 $ forces $ a = b = c = d = 4, $ so \[P(z) = 4z^4 + 4z^3 + 4z^2 + 4z + 4 = 4(z^4 + z^3 + z^2 + z + 1) = 0.\]If $ z^4 + z^3 + z^2 + z + 1 = 0, $ then \[(z - 1)(z^4 + z^3 + z^2 + z + 1) = 0,\]which becomes $ z^5 - 1 = 0 $. Then $ z^5 = 1, $ so $ |z^5| = 1 $. Hence, $ |z|^5 = 1, $ so $ |z| = 1 $. This confirms that all the roots of $ z^4 + z^3 + z^2 + z + 1 $ have magnitude 1, and $ P(1) = 20 $. Otherwise, we can assume that $ d \neq 4 $. Then from equation $ (*), $ \[2x_0 = \frac{a - c}{d - 4}.\]Multiplying the equation $ p - 8x_0 = a $ by $ d, $ we get \[dp - 8dx_0 = ad.\]Multiplying the equation $ p - 2dx_0 = c $ by 4, we get \[4p - 8dx_0 = 4c.\]Subtracting these equations, we get $ dp - 4p = ad - 4c, $ so \[p = \frac{ad - 4c}{d - 4}.\]Let \[k = 2px_0 = 2x_0 \cdot p = \frac{a - c}{d - 4} \cdot \frac{ad - 4c}{d - 4} = \frac{(a - c)(ad - 4c)}{(d - 4)^2}.\]Then from the equation $ d - 2px_0 + 4 = b, $ $ k = d - b + 4 $. Since $ b \le 4, $ $ k \ge 0 $. We then divide into the cases where $ a = c $ and $ a > c $. Case 1: $ a=c $. In this case, $ k=0 $ and $ b=d+4 $, so $ a=b=c=4 $ and $ d=0 $. We have already covered these possibilities when we looked at the case where $ z = -1 $ was a root of $ P(z) $. Case 2: $ a>c\geq 0 $. Since $ k\geq 0 $, we have $ ad-4c\geq 0, $ or $ ad \ge 4c $. However, $ ad \leq 4c $, so $ ad = 4c $. For this to hold, we must have $ c = d $. Then we obtain $ k=0 $ again. In this case, $ b=d+4 $, so $ a=b=4 $ and $ c=d=0, $ and \[P(z) = 4z^4 + 4z^3 + 4z^2 = 4z^2 (z^2 + z + 1).\]The roots of $ z^2 + z + 1 = 0 $ are $ z = -\frac{1}{2} \pm \frac{\sqrt{3}}{2} i, $ which have magnitude 1, and $ P(1) = 12 $. Therefore, the desired sum is $ 60 + 20 + 12 = \boxed{92} $.