A bee starts flying from point $P_0$. She flies 1 inch due east to point $P_1$. For $j \ge 1$, once the bee reaches point $P_j$, she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$. When the bee reaches $P_{2015}$ she is exactly $a\sqrt{b} + c\sqrt{d}$ inches away from $P_0$, where $a$, $b$, $c$ and $d$ are positive integers and $b$ and $d$ are not divisible by the square of any prime. What is $a+b+c+d$?

$ \textbf{(A)}\ 2016 \qquad\textbf{(B)}\ 2024 \qquad\textbf{(C)}\ 2032 \qquad\textbf{(D)}\ 2040 \qquad\textbf{(E)}\ 2048$
1. Define \(\omega = e^{\frac{i\pi}{6}}\). Note that \(P_{2015}\) is the sum of the points \(1\omega^0, 2\omega^1, \ldots, 2015\omega^{2014}\) by translating each move to the origin. Hence, we want to evaluate \(1\omega^0 + 2\omega^1 + \ldots + 2015\omega^{2014}\).

2. Consider the polynomial \(P(x) = \frac{x^{2016} - 1}{x - 1} = x^{2015} + x^{2014} + \ldots + 1\). Then,
   \[
   P'(x) = \frac{d}{dx} \left( \frac{x^{2016} - 1}{x - 1} \right)
   \]
   Using the quotient rule, we get:
   \[
   P'(x) = \frac{(2016x^{2015})(x - 1) - (x^{2016} - 1)}{(x - 1)^2} = \frac{2016x^{2015}(x - 1) - (x^{2016} - 1)}{(x - 1)^2}
   \]
   Simplifying the numerator:
   \[
   P'(x) = \frac{2016x^{2015}x - 2016x^{2015} - x^{2016} + 1}{(x - 1)^2} = \frac{2015x^{2016} - 2016x^{2015} + 1}{(x - 1)^2}
   \]
   This simplifies to:
   \[
   P'(x) = 2015x^{2014} + 2014x^{2013} + \ldots + 1
   \]

3. We seek \(P'(\omega)\). Since \(x^{12} = 1\), we can write:
   \[
   P'(\omega) = \frac{2016 - \frac{2016}{\omega}}{(\omega - 1)^2} = \frac{2016}{\omega(\omega - 1)} = \frac{2016}{\omega^2 - \omega}
   \]

4. Now, \(\omega = \frac{\sqrt{3}}{2} + \frac{1}{2}i\) and \(\omega^2 = \frac{1}{2} + \frac{\sqrt{3}}{2}i\). Hence,
   \[
   \omega^2 - \omega = \left( \frac{1}{2} + \frac{\sqrt{3}}{2}i \right) - \left( \frac{\sqrt{3}}{2} + \frac{1}{2}i \right) = \frac{1}{2} - \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}i - \frac{1}{2}i
   \]
   Simplifying further:
   \[
   \omega^2 - \omega = \frac{1 - \sqrt{3}}{2} + \frac{\sqrt{3} - 1}{2}i
   \]

5. Therefore,
   \[
   P'(\omega) = \frac{2016}{\frac{1 - \sqrt{3}}{2} + \frac{\sqrt{3} - 1}{2}i} = \frac{2016 \cdot 2}{(1 - \sqrt{3}) + (\sqrt{3} - 1)i} = \frac{4032}{1 - \sqrt{3} + (\sqrt{3} - 1)i}
   \]

6. To simplify, multiply the numerator and the denominator by the conjugate of the denominator:
   \[
   P'(\omega) = \frac{4032 (1 + \sqrt{3} - (\sqrt{3} - 1)i)}{(1 - \sqrt{3})^2 - (\sqrt{3} - 1)^2}
   \]
   Simplifying the denominator:
   \[
   (1 - \sqrt{3})^2 - (\sqrt{3} - 1)^2 = (1 - \sqrt{3})^2 - (\sqrt{3} - 1)^2 = 1 - 2\sqrt{3} + 3 - (3 - 2\sqrt{3} + 1) = 4 - 4 = 0
   \]
   This indicates a mistake in the simplification process. Instead, we should directly use the magnitude approach.

7. The magnitude of \(P'(\omega)\) is:
   \[
   |P'(\omega)| = \left| \frac{2016}{\omega^2 - \omega} \right| = \frac{2016}{|\omega^2 - \omega|}
   \]
   Since \(|\omega| = 1\), we have:
   \[
   |\omega^2 - \omega| = \left| \frac{1 - \sqrt{3}}{2} + \frac{\sqrt{3} - 1}{2}i \right| = \sqrt{\left( \frac{1 - \sqrt{3}}{2} \right)^2 + \left( \frac{\sqrt{3} - 1}{2}i \right)^2} = \sqrt{\frac{(1 - \sqrt{3})^2 + (\sqrt{3} - 1)^2}{4}}
   \]
   Simplifying:
   \[
   = \sqrt{\frac{1 - 2\sqrt{3} + 3 + 3 - 2\sqrt{3} + 1}{4}} = \sqrt{\frac{8 - 4\sqrt{3}}{4}} = \sqrt{2 - \sqrt{3}}
   \]

8. Therefore,
   \[
   |P'(\omega)| = \frac{2016}{\sqrt{2 - \sqrt{3}}}
   \]

9. Finally, we need to find the value of \(a + b + c + d\) where \(a\sqrt{b} + c\sqrt{d}\) represents the distance. From the simplification, we get:
   \[
   1008\sqrt{2} + 1008\sqrt{6}
   \]
   Thus, \(a = 1008\), \(b = 2\), \(c = 1008\), and \(d = 6\). Therefore,
   \[
   a + b + c + d = 1008 + 2 + 1008 + 6 = 2024
   \]

The final answer is \(\boxed{2024}\)