Answer
$ B=\dfrac{\mu_0 I \theta}{4\pi R } $
Work Step by Step
We have here 3 wires, the two straight wires plus the circular arc.
The magnitudes of the magnetic fields due to the two straight wires, at point P, are zeros since the angles are zeros.
Now it is obvious that the net magnetic field at point P is due to the circular arc only which is a part of a loop.
Recalling that the magnitude of the magnetic field of a very short segment $ds$ is given by
$$dB=\dfrac{\mu_0 I ds}{4\pi R^2} $$
where for a small segment of the arc $ds=Rd\theta$ where $\theta$ in radians.
So,
$$dB=\dfrac{\mu_0 I Rd\theta}{4\pi R^2} =\dfrac{\mu_0 I d\theta}{4\pi R } $$
Taking the integral,
$$\int_0^BdB=\int_0^\theta\dfrac{\mu_0 I d\theta}{4\pi R } $$
$$ B=\dfrac{\mu_0 I }{4\pi R } \int_0^\theta d\theta $$
$$ \boxed{B=\dfrac{\mu_0 I \theta}{4\pi R } }$$
