Answer
$Kp=1.483\times10^{3}$
Products are favored at equilibrium.
Work Step by Step
It is observed,all the coefficients of that reaction were divided by 2 (or multiplied by $\frac{1}{2})$. Thus, the new equilibrium constant is equal to the old one to the power of $\frac{1}{2}$
$Kp=(2.2\times10^{6})^{\frac{1}{2}}$=$1.483\times10^{3}$
Since $Kp\gt1$,the products will be favored at equilibrium.
