$\mathrm{B}_{\mathrm{AB}}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I}}{\mathrm{r}} \otimes$

The magnetic field at point $\mathrm{P}$ due to current $I$ in $BC$ is, $\mathrm{B}_{\mathrm{BC}}=0$ (As the point $\mathrm{P}$ is along the $BC$) The magnetic field at point $P$ due to current $I$ in $A'B'$ is.

${B_{A'B'}} = \frac{{{\mu _0}I}}{{4\pi r}}\otimes $

The magnetic field at point $\mathrm{P}$ due to current $I$ in $B^{\prime} C^{\prime}$ is

$\left.\mathrm{B}_{\mathrm{BC}}=0 \quad \text { (As the point } \mathrm{P} \text { is along the BC }\right)$

$\therefore $  The net magnetic field at $\mathrm{P}$ is,

$\mathrm{B}=\mathrm{B}_{\mathrm{AB}}+\mathrm{B}_{\mathrm{BC}}+\mathrm{B}_{\mathrm{AB}}+\mathrm{B}_{\mathrm{BC}}$

$\frac{\mu_{0} I}{4 \pi r}+0+\frac{\mu_{0} I}{4 \pi r}+0=2\left(\frac{\mu_{0} I}{4 \pi r}\right)=\frac{\mu_{0}}{4 \pi}\left(\frac{2 I}{r}\right)$