We know that x can only take values 0,1,2,3 since P(X=x)=0 for everything else.
P(X=0)=2k
P(X=1)=k(1)=k
P(X=2)=k(2−1)=k
P(X=3)=k(3−1)=2k
For any probability distribution, the sum of all probabilities must always equal 1 (since one of the outcomes has to happen).
P(X=0)+P(X=1)+P(X=2)+P(X=3)=1
2k+k+k+2k=1
6k=1
k=61