In given case,

$d=R \sin 45^{\circ}=\frac{R}{\sqrt{2}}$

$\theta_{1}=135^{\circ}, \theta_{2}=180^{\circ}$

$\therefore B($ at $P)=\frac{\mu_{0} I}{4 \pi\left(\frac{R}{\sqrt{2}}\right)}\left[\cos 135^{\circ}-\cos 180^{\circ}\right]$

$=\frac{\mu_{0} I}{4 \pi R} \sqrt{2}\left(\frac{-1}{\sqrt{2}}-(-1)\right)$

$=\frac{\mu_{0} I}{4 \pi R} \sqrt{2}\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)$

or $B($ at $P)=\frac{\mu_{\rho} I}{4 \pi R}(\sqrt{2}-1) T$