Reaction stoichiometry for calculation of Kp NH$_2$CO$_2$NH$_4$(s)⇌$2$ NH$_3$(g)+ CO$_2$(g)

I want to calculate K$_p$ for the reaction above, given that the total pressure is $0.63$ bar.
My calculations:
$0.63$ = p(CO$_2$) + p(NH$_3$)
,   where p(NH$_3$) = $1/2$ p(CO$_2$) (For every NH$_3$ half a CO$_2$ is needed)
$0.63$ = p(CO$_2$) + ($1/2$ p(CO$_2$) = $3/2$p(CO$_2$) -&gt; p(CO$_2$) = $0.42$ bar , p(NH$_3$) = $0.21$ bar
So, K$_p$ = ($0.21$)$^2$ * ($0.42$) = $0.0185$ bar$^3$
However, the answer key's way of calculating this is:
$0.63$ = p(CO$_2$) + p(NH$_3$)
,   where p(NH$_3$) = $2$ p(CO$_2$)
$0.63$ = $3$p(CO$_2$) , so that  p(CO$_2$) = $0.21$ bar and p(NH$_3$) = $0.42$ bar
How can p(NH$_3$) = $2$ p(CO$_2$)? Isn't that wrong? Or am I thinking wrong? I have contacted my professor but haven't received an answer yet and our exam is tomorrow so hopefully someone here can help me understand. Thank you!
Each time one molecule of salt is decomposed, it produced $\ce{2 NH3}$ for $\ce{1 CO2}$. The number of molecules $\ce{NH3}$ is always double from that of $\ce{CO2}$ in the gas phase. Now the partial pressure of a given compound is proportional to the number of molecules of this compound in the gas phase. As a consequence, the partial pressure of $\ce{NH3}$ is always twice as high as the partial pressure of $\ce{CO2}$