1. Electric field due to a dipole on axial and equatorial lines

For an ideal dipole of moment $p$ at a point at distance $r$ from its centre:

• On the axial line, $E_{\text{axial}}=\frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}$
• On the equatorial line, $E_{\text{equatorial}}=\frac{1}{4\pi\varepsilon_0}\frac{p}{r^3}$

2. Interpret the figure

Since point $C$ lies on the axis of dipole $A$, the field due to dipole $A$ at $C$ is along the axis of dipole $A$.

From the usual configuration implied by the question, dipole $B$ is perpendicular to dipole $A$, and point $C$ lies on the equatorial line of dipole $B$.

Let the distance $OC=r$.

Then,

• Field at $C$ due to dipole $A$: $E_A=\frac{1}{4\pi\varepsilon_0}\frac{2p_1}{r^3}$
• Field at $C$ due to dipole $B$: $E_B=\frac{1}{4\pi\varepsilon_0}\frac{p_2}{r^3}$

These two fields are perpendicular.

3. Use the given angle

The resultant electric field makes an angle $37^\circ$ with the axis of dipole $A$.

So, $\tan 37^\circ=\frac{E_B}{E_A}$

Given, $\sin 37^\circ=\frac{3}{5}$ which implies $\cos 37^\circ=\frac{4}{5}$ Hence, $\tan 37^\circ=\frac{3}{4}$

Therefore, $\frac{E_B}{E_A}=\frac{3}{4}$

Substitute the expressions: $\frac{\frac{1}{4\pi\varepsilon_0}\frac{p_2}{r^3}}{\frac{1}{4\pi\varepsilon_0}\frac{2p_1}{r^3}}=\frac{3}{4}$

$\frac{p_2}{2p_1}=\frac{3}{4}$

$4p_2=6p_1$

$\frac{p_1}{p_2}=\frac{2}{3}$

4. Match with options

$\boxed{\frac{p_1}{p_2}=\frac{2}{3}}$

So the correct option is C.