Answer: 19

Solution:

Identify the plane with the complex plane, so a move sends $z$ to $\omega z + 10$ with $\omega = e^{i\pi/4}.$ Starting from $z_0 = 5$ and iterating, $$\begin{aligned} z_{150} &= 5\omega^{150} \\ &\quad {}+ 10 \scriptsize\left(\omega^{149} + \omega^{148} + \cdots + \omega + 1\right). \end{aligned}$$

Since $\omega^8 = 1$ and $150 = 8 \cdot 18 + 6,$ we get $\omega^{150} = \omega^6 = -i.$ In the geometric sum, every block of $8$ consecutive powers adds to $0,$ so the $150$ terms reduce to $1 + \omega + \cdots + \omega^5$ $= -\omega^6 - \omega^7$ $= i - \left(\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i\right).$ Therefore $$\begin{aligned} z_{150} &= -5i \\ &\quad {}+ 10 \scriptsize\left(-\frac{\sqrt{2}}{2} + \left(1 + \frac{\sqrt{2}}{2}\right)i\right) \\ &= -5\sqrt{2} + \left(5 + 5\sqrt{2}\right)i. \end{aligned}$$

Thus $|p| + |q| = 5\sqrt{2} + 5 + 5\sqrt{2}$ $= 5 + 10\sqrt{2} \approx 19.14,$ and the greatest integer less than or equal to this is $19.$