$K_p=0.11Atm^2$
Start by writing down the system which is at equilibrium:
$NH_4HS_((s))rightleftharpoonsNH_(3(g))+H_2S_((g))$
The solid reactant has zero partial pressure
So $K_p$= $P_(NH_(3)).P_(H_2S)$
Where $P$ denotes the of each gas.
$P_(NH_(3))+P_(H_2S)=0.66Atm$
So $P_(NH_(3))=0.33Atm$
and $P_(H_2S)=0.33Atm$
since they are formed in a 1:1 ratio
So
$K_p=(0.33)^2=0.11Atm^2$