Answer
The net magnetic field 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 $R = 5.0~cm$:
$B = \frac{\mu_0~i_1~\phi}{4~\pi~R}$
$B = \frac{(4\pi\times 10^{-7}~H/m)~(0.40~A)~(\pi~rad)}{(4~\pi)~(0.050~m)}$
$B = 2.51\times 10^{-6}~T$
By the right hand rule, this magnetic field is into the page.
We can find the magnetic field at $P$ due to the arc of current with radius $R = 4.0~cm$:
$B = \frac{\mu_0~i_2~\phi}{4~\pi~R}$
$B = \frac{(4\pi\times 10^{-7}~H/m)~(0.80~A)~(2\pi/3~rad)}{(4~\pi)~(0.040~m)}$
$B = 4.19\times 10^{-6}~T$
By the right hand rule, this magnetic field is into the page.
Since the magnetic field due to each arc of current is into the page, the net magnetic field is into the page.
