Correct Answer  Option 1 :
momentum of 1^{st} body > momentum of 2^{nd} body
The correct answer is option 1) i.e. momentum of 1^{st} body > momentum of 2^{nd} body
CONCEPT:

Kinetic energy (KE): The energy due to the motion of the body is called kinetic energy.
KE = (1/2) m v^{2}

Momentum (p): The product of mass and velocity is called momentum.
p = m v
Where m is mass and v is velocity
EXPLANATION:
K_{1} = (1/2) m_{1} v_{1}^{2}
K_{2} = (1/2) m_{2} v_{2}^{2}
Given that:
The kinetic energies of objects A and B are equal.
K_{1} = K_{2}
The momenta of objects A and B:
p_{1} = m_{1 }× v_{1} and
p_{2} = m_{2} × v_{2}
We know that
v_{1} < v_{2}
Divide the numerator and denominator in the above by K_{1} and K_{2} (note K_{1} = K_{2}), to obtain
v_{1}/K_{1} < v_{2}/K_{2}
Which gives
K_{1}/v_{1} > K_{2}/v_{2}
Substitute K_{1} and K_{2} by their expressions given above
(1/2) m_{1} v_{1}^{2} / v_{1} > (1/2) m_{2} v_{2}^{2} /v_{2}
Simplify to obtain
m_{1}v_{1} > m_{2 }v_{2}
Which gives p_{1} > p_{2}