How to determine the total pressure of a gaseous mixture of xenon tetrafluoride and hydrogen?
You are correct up to the last part. The total pressure would be the sum of all the partial pressures in the reaction vessel. 

We know that $$P=\left(\frac{RT}{V}\right)n$$ and therefore $n$ varies directly with $P$ as $T$, $R$, and $V$ are constant. For the sake of the problem (it won't matter) we can just say that $x=\left(\frac{RT}{V}\right)$. Therefore, $P=xn$ and the starting amounts of substance are equal to $\frac{2}{x}\pu{ mol}$ $\ce{XeF4}$ and $\frac{10}{x}\pu{ mol}$ $\ce{H2}$. 

From the balanced equation and the starting materials and stoichiometry, we know that $\frac{2}{x}\pu{ mol}$ $\ce{XeF4}$ would react with $\frac{4}{x}\pu{ mol}$ $\ce{H2}$ to form $\frac{2}{x}\pu{ mol}$ $\ce{Xe}$ and $\frac{8}{x}\pu{ mol}$  $\ce{HF}$ to make $\frac{10}{x}\pu{ mol}$ product. 

However, you only reacted $\frac{4}{x}\pu{ mol}$ $\ce{H2}$ of the $\frac{10}{x}\pu{ mol}$ total, so there is $\frac{6}{x}\pu{ mol}$ unreacted $\ce{H2}$ left in the reaction vessel. 

$$\begin{multline}
  \frac{10}{x}\text{ amount of substance of product } 
+ \frac{6}{x}\text{ amount of substance of unreacted }\\
= \frac{16}{x}\text{ total amount of substance}
\end{multline}$$

From $P=xn$, the fact that this occurred instantly, and $T$, $R$, and $V$ are constant, we then know that the final pressure is $\ce{16 kPa}$.