Answer
The net magnetic field at $P$ is into the page.
Work Step by Step
Note that the straight sections of wire do not contribute to the magnetic field at $P$
We can write the expression for the magnetic field due to an arc of current:
$B = \frac{\mu_0~i~\phi}{4~\pi~R}$
We can find the magnetic field at $P$ due to the arc of current with radius $r = 4.00~m$:
$B = \frac{\mu_0~i~\phi}{4~\pi~r}$
$B = \frac{(4\pi\times 10^{-7}~H/m)~(0.200~A)~(7\pi/4~rad)}{(4~\pi)~(4.00~m)}$
$B = 2.75\times 10^{-8}~T$
By the right hand rule, this magnetic field is out of the page.
We can find the magnetic field at $P$ due to the arc of current with radius $r = 2.00~m$:
$B = \frac{\mu_0~i~\phi}{4~\pi~r}$
$B = \frac{(4\pi\times 10^{-7}~H/m)~(0.200~A)~(7\pi/4~rad)}{(4~\pi)~(2.00~m)}$
$B = 5.50\times 10^{-8}~T$
By the right hand rule, this magnetic field is into the page.
Since the magnitude of the magnetic field into the page is greater than the magnitude of the magnetic field out of the page, the net magnetic field at $P$ is into the page.
