Answer
$p_{f1}= (2\;\hat i+4\;\hat j)\;\rm kg\cdot m/s $
Work Step by Step
We can assume that the collision is perfectly elastic and the system (the two particles) is isolated.
This means that the momentum is conserved.
Thus,
$$p_i=p_f$$
Thus,
$$p_{i1}+p_{i2}=p_{f1}+p_{f2}\tag 1$$
We can see, from the given graph, that
$$p_{i1}=(-2\;\hat i+2\;\hat j)\;\rm kg\cdot m/s$$
$$p_{i2}=(4\;\hat i+1\;\hat j)\;\rm kg\cdot m/s$$
$$p_{f2}=(0\;\hat i-1\;\hat j)\;\rm kg\cdot m/s$$
Plugging these last 3 equations into (1);
$$-2\;\hat i+2\;\hat j+4\;\hat i+1\;\hat j=p_{f1}+0\;\hat i-1\;\hat j$$
Solving for $p_{f1}$
$$p_{f1}=-2\;\hat i+2\;\hat j+4\;\hat i+1\;\hat j-0\;\hat i+1\;\hat j $$
$$p_{f1}= (\color{red}{\bf2}\;\hat i+\color{red}{4}\;\hat j)\;\rm kg\cdot m/s $$
