Calculating partial pressure of carbon monoxide
For simplicity, let's assign numerical indices to the compounds of interest — all gaseous products participating in equilibrium:
$$\ce{ZnO(s) + \underset{1}{CO(g)} &lt;=&gt; \underset{2}{Zn(g)} + \underset{3}{CO2(g)}}$$
Partial pressure of carbon monoxide can be found via its mole fraction $x_1$ and given total pressure $p$:
$$p_1 = x_1p\tag{1}$$
To find unknown mole fraction, two facts can be taken into consideration:

As Karsten also pointed out, the molar ratio between the products is $1:1$ so that $x_2 = x_3$. Which gives
$$x_1 + x_2 + x_3 = 1 \quad\implies\quad x_2 = x_3 = \frac{1 - x_1}{2}\label{eqn:2}\tag{2}$$

All three mole fractions are linked together in the expression for the average molar mass:
$$\bar{M} = x_1M_1 + x_2M_2 + x_3M_3\label{eqn:3}\tag{3}$$


The average molar mass can be found via average density of the gaseous mix $\bar{ρ}$ by applying ideal gas law:
$$pV = nRT$$
$$p = \frac{nRT}{V} = \frac{mRT}{\bar{M}V} = \frac{\bar{ρ}RT}{\bar{M}}$$
$$
\begin{align}
\bar{M} &amp;= \frac{\bar{ρ}RT}{p} \\
  &amp;= \frac{\pu{0.344 g L-1} × \pu{0.082 L atm K-1 mol-1} × \pu{1300 K}}{\pu{1 atm}} \\
  &amp;\approx \pu{36.67 g mol-1}\tag{4}
\end{align}
$$
Now, knowing numerical value for $\bar{M},$ we can combine \eqref{eqn:2} and \eqref{eqn:3}:
$$\bar{M} = x_1M_1 + \frac{1 - x_1}{2}M_2 + \frac{1 - x_1}{2}M_3$$
Grouping the like terms and expressing $x_1$:
$$
\begin{align}
x_1 &amp;= \frac{M_2 + M_3 - 2\bar{M}}{M_2 + M_3 - 2M_1} \\
  &amp;= \frac{\pu{44.01 g mol-1} + \pu{65.38 g mol-1} -2×\pu{36.67 g mol-1}}{\pu{44.01 g mol-1} + \pu{65.38 g mol-1} -2×\pu{28.01 g mol-1}} \\
  &amp;\approx 0.67\tag{5}
\end{align}
$$
Finally, the partial pressure of carbon monoxide:
$$p_1 = 0.67×\pu{1 atm} = \pu{0.67 atm}$$