Answer
$p_A \gt p_B$
Work Step by Step
We can write an expression for the momentum of particle A:
$p_A = \gamma_A~m_A~v_a$
We can write an expression for the momentum of particle B:
$p_B = \gamma_B~m_B~v_B$
$p_B = \gamma_B~(2m_A)~(v_A/2)$
$p_B = \gamma_B~m_A~v_A$
In general:
$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$
As $v$ increases, the value of $\gamma$ also increases.
Therefore, $\gamma_A \gt \gamma_B$, since $v_A \gt v_B$
Then, $p_A \gt p_B$, since $\gamma_A \gt \gamma_B$