As illustrated in the figure, two short magnets with equal dipole...

Let the centre of the two magnets be at the origin $(0,0)$.

The horizontal magnet has its South pole on the left and North pole on the right. Its magnetic dipole moment is directed from South to North, so $\vec{M}_1 = M \hat{i}$.

The vertical magnet has its South pole at the bottom and North pole at the top. Its magnetic dipole moment is also directed from South to North, so $\vec{M}_2 = M \hat{j}$.

The net magnetic dipole moment of the system is the vector sum of the individual dipole moments:
$\vec{M}_{net} = \vec{M}_1 + \vec{M}_2 = M \hat{i} + M \hat{j}$

The magnitude of the net dipole moment is:
$|\vec{M}_{net}| = \sqrt{M^2 + M^2} = \sqrt{2}M$

The direction of $\vec{M}_{net}$ is at an angle of $45^{\circ}$ to the positive x-axis, which is along the bisector of the angle between the two North poles.

The point $P$ is at a distance $d$ along the bisector of the angle between the two South poles. This direction is exactly opposite to the direction of the net dipole moment. Therefore, the point $P$ lies on the axial line of the equivalent net dipole.

The magnitude of the magnetic field on the axial line of a short dipole at a distance $d$ is given by:
$B = \dfrac{\mu_0}{4\pi} \dfrac{2|\vec{M}_{net}|}{d^3}$

Substituting $|\vec{M}_{net}| = \sqrt{2}M$, we get:
$B = \dfrac{\mu_0}{4\pi} \dfrac{2\sqrt{2}M}{d^3}$