Step 1: Field due to the first magnet.
Point P is on the axis (axial position) of the first magnet with moment \(M\).
For a short magnet, the axial field is:
\[ B_{axial} = \frac{\mu_0}{4\pi} \frac{2M}{d^3} \]
Given that this value is \(B\). So, \( B = \frac{\mu_0}{4\pi} \frac{2M}{d^3} \).
Step 2: Field due to the second magnet.
The second magnet has moment \(2M\) and is perpendicular to the first. Since their centers coincide and P lies on the axis of the first magnet, P lies on the equatorial line of the second magnet.
The equatorial field is given by:
\[ B_{equatorial} = \frac{\mu_0}{4\pi} \frac{M'}{d^3} \]
Here \(M' = 2M\).
\[ B' = \frac{\mu_0}{4\pi} \frac{2M}{d^3} \]
Notice that this expression is exactly equal to \(B\) (the field of the first magnet).
So, \( B' = B \).
Step 3: Resultant Field.
The axial field \(B\) is along the magnetic moment vector. The equatorial field \(B'\) is opposite to the magnetic moment vector of the second magnet. Since the two magnets are perpendicular, the two magnetic field vectors at P are perpendicular to each other.
\[ B_{net} = \sqrt{B^2 + (B')^2} = \sqrt{B^2 + B^2} = \sqrt{2}B \]