Answer
Point P lies between the third minimum and the third maximum, but closer to the third maximum.
Work Step by Step
We can find the path length difference as a multiple of $\lambda$:
$\Delta L = d~sin~\theta$
$\Delta L = (4.24\times 10^{-6}~m)~sin~20.0^{\circ}$
$\Delta L = 1.45\times 10^{-6}~m$
We can find the phase difference as a multiple of $\lambda$:
$\frac{\Delta L}{\lambda} = \frac{1.45\times 10^{-6}~m}{500\times 10^{-9}~m} = 2.9$
The phase difference is $~~2.9$
The third minimum corresponds to a phase difference of $2.5$
The third maximum corresponds to a phase difference of $3.0$
Therefore, point P lies between the third minimum and the third maximum, but closer to the third maximum.
