To solve the given problem, we need to understand the concept of how two prisms are combined to produce dispersion without deviation. This occurs when the angle of deviation due to one prism cancels out the angle of deviation due to the other prism.

The conditions for the elimination of deviation while maintaining dispersion is defined by the relationship:

\((\mu_1 - 1)A_1 = (\mu_2 - 1)A_2\)

where:

• \(\mu_1\) = Refractive index of prism \(P_1\)
• \(A_1\) = Angle of prism \(P_1\)
• \(\mu_2\) = Refractive index of prism \(P_2\)
• \(A_2\) = Angle of prism \(P_2\)

Given:

• \(\mu_1 = 1.54\)
• \(A_1 = 4^\circ\)
• \(\mu_2 = 1.72\)
• We need to find \(A_2\).

Substitute the given values into the equation:

\((1.54 - 1) \times 4 = (1.72 - 1) \times A_2\)

This simplifies to:

\(0.54 \times 4 = 0.72 \times A_2\)

Calculate the left-hand side:

\(2.16 = 0.72 \times A_2\)

Now, solve for \(A_2\):

\(A_2 = \frac{2.16}{0.72} = 3^\circ\)

Therefore, the angle of prism \(P_2\) must be \(3^\circ\) to achieve dispersion without deviation. Hence, the correct option is 3°.