Answer
The phase difference at $P_2$ is $~~3.5$
Work Step by Step
We can find the path length difference to $P_2$:
$\Delta L = 1360~nm-80~nm = 1280~nm$
We can find the phase difference at $P_2$ if the waves started out in phase:
$(\frac{1280~nm}{400~nm})~(2\pi~rad) = 6.4~\pi~rad$
However, the waves did not start out in phase. Since the path length difference to $P_1$ is zero, the wave from $S_2$ starts out $0.600\pi~rad$ ahead of the wave from $S_1$
We can find the total phase difference at $P_2$:
$6.4\pi~rad+0.600\pi~rad = 7.0\pi~rad = 3.5~(2\pi)~rad$
As a multiple of $\lambda$, the phase difference at $P_2$ is $~~3.5$
