Hello Aspirant!
We know that -
Momentum (p) = m x v
where, m=mass of the object
v= velocity of the object
=> v = p/m
Now, Kinetic Energy (K. E.) = (1/2) x m x (v)^2
Substituting the value of v = p/m in this equation, we get -
K. E. = (1/2) x m x (p/m)^2
= (1/2) x (1/m) (p^2)
According to the question, for 20% increase in momentum, kinetic energy will be -

K. E.' = (1/2) x (1/m) x [p+ (20p/100)]^2
= (1/2) x (1/m) x [p^2 + [2 x p x (20p/100)] + (20p/100)^2]

{Using identity (a+b)^2 = a^2 + (2 x a x b) + b^2 and here : a= p, b= (20p/100)}

= (1/2) x (1/m) x [p^2 + ((2/5) x p^2) + ((400/10000) x p^2)]
= (1/2) x (1/m) x (p^2)[1+ (2/5) + (1/25)]
{Taking (p^2) out as common}
= (1/2) x (1/m) x (p^2)[(25+10+1)/25]
= (1/2) x (1/m) x p^2 [36/25]
So, K. E.' = {(1/2) x (1/m) x (p^2)} [1.44]
Or K.E.' = {K.E.} x [1.44]

So, Change in Kinetic energy :
K.E.' - K. E. = 1.44 K.E. - K. E.
Change in K.E. = 0.44
Thus, (Change in K.E.) x 100 = 44%
That means, for 20% increase in Momentum, the Kinetic energy increases by 44%.

Hope this answer will help you.