Answer
$B_{net} = 1.7\times 10^{-6}~T$
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 out of 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.
We can find the net magnetic field at $P$:
$B_{net} = (4.19\times 10^{-6}~T)-(2.51\times 10^{-6}~T)$
$B_{net} = 1.7\times 10^{-6}~T$
