1. Interpret the hairpin-shaped wire

   The wire consists of:

   • two straight semi-infinite segments,
   • joined by a semicircular arc of radius $r$.

   Point $P$ is the center of the semicircle.

2. Magnetic field due to the semicircular arc

   For a circular arc of angle $\theta$ at the center, $B_{\text{arc}}=\frac{\mu_0 I\theta}{4\pi r}$

   Here $\theta=\pi$ for a semicircle, so $B_{\text{semi}}=\frac{\mu_0 I\pi}{4\pi r}=\frac{\mu_0 I}{4r}$

   Writing with denominator $4\pi r$, $B_{\text{semi}}=\frac{\mu_0 I}{4\pi r}\,\pi$

3. Magnetic field due to each straight semi-infinite part

   The perpendicular distance of point $P$ from each straight segment is $r$.

   For a semi-infinite straight wire, $B_{\text{semi-inf}}=\frac{\mu_0 I}{4\pi r}$

   Since there are two such straight parts, $B_{\text{straight total}}=2\cdot \frac{\mu_0 I}{4\pi r}=\frac{\mu_0 I}{4\pi r}\cdot 2$

4. Direction of fields

   Using the right-hand rule, the magnetic field at $P$ due to the semicircular arc and due to both straight segments is in the same direction, so magnitudes add.

5. Net magnetic field

   $B= B_{\text{semi}}+B_{\text{straight total}}$ $B=\frac{\mu_0 I}{4\pi r}(\pi+2)$

   Hence, $\boxed{B=\frac{\mu_0 I}{4\pi r}(2+\pi)}$

6. Option check

   • A: $\frac{\mu_0 I}{4\pi r}(2-\pi)$ — incorrect
   • B: $\frac{\mu_0 I}{2\pi r}(2-\pi)$ — incorrect
   • C: $\frac{\mu_0 I}{4\pi r}(2+\pi)$ — correct
   • D: $\frac{\mu_0 I}{2\pi r}(2+\pi)$ — incorrect

Therefore, the correct option is C.