Answer
$$B={\mu_0 I }/{4 R }$$
Work Step by Step
We have here 3 wires, the two straight wires plus the semicircular wire.
The magnitudes of the magnetic fields due to the two straight wires, at the center point of the semicircular wire, 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,
$$\int_0^BdB=\int_0^\theta\dfrac{\mu_0 I d\theta}{4\pi R } $$
We can use $\theta$ from $-\pi /2$ to $\pi /2$, or as seen below.
$$ B=\dfrac{\mu_0 I }{4\pi R } \int_0^\pi d\theta $$
$$B=\dfrac{\mu_0 I \pi}{4\pi R } $$
$$ \boxed{B=\dfrac{\mu_0 I }{4 R } }$$