What is the percent change in momentum of a proton that accelerates from 0.41c to 0.85c?

At these speeds, we can't ignore the relativistic masses.

At 0.41c: m = m₀ / √ (1-.41²) = m₀ / √0.8319 = m₀ / 0.9121 = 1.096 m₀

At 0.85c: m = m₀ / √ (1-.85²) = m₀ / √0.2775 = m₀ / 0.5268 = 1.898 m₀

The change in momentum is (1.898m₀ x 0.85c) - (1.096m₀ x 0.41c)

= (1.613 - 0.449) m₀ c = 1.164 m₀ c.

As a fraction of the original momentum, that's (1.164 / 0.449)

= (2.592) = 159% increase.