Finding initial pressure in a chemical equilibrium
If the initial pressure of $\ce{CO}$ and $\ce{H_2O}$ are equal, then let $p_{\ce{H_2O}}=p_{\ce{CO}}=x$.
Since their total is 1.5 bar, i.e. $\ce{p_{H_2O} + p_{CO}}=1.5$
Can you solve this equation?

Edit: This is under the assumption that there was originally no carbon dioxide or hydrogen, I have not yet managed to generalise it to any case.
It is given that $K=4$. From the equation for the equilibrium constant, $K=\frac{\ce{[CO2][H2]}}{\ce{[CO][H2O]}}$.
It is also given that initially, $\ce{p_{H2O} = p_{\ce{CO}}}$; from the ideal gas law,
$PV=nRT \implies P=\frac{nRT}{V}=kC  \tag{1}$
where $k$ is some constant and C is concentration.
So initially, $[\ce{H2O}]=[\ce{CO}]=y$, where $y$ is a constant. Let the concentration consumed by the equilibrium shifting to the right be x.
At equilibrium, $\frac{\ce{[CO2][H_2]}}{[\ce{CO}][\ce{H2O}]}=\frac{x^2}{(y-x)^2}=4$.
Solving this equation gives $x=\frac{2y}{3}$.
Now then, if you look back at (1), pressure is proportional to concentration. Try to use that relationship to solve the problem.

 Now, from $(1)$, we see that pressure is directly proportional to concentration. We have worked out from our algebra that: $[\ce{CO2}]=[\ce{H2}]=\frac{2y}{3}$, therefore $[\ce{CO}]=[\ce{H2O}]=\frac{y}{3}$.

So then $p_{\ce{H2}}= p_{\ce{CO2}} = 2p_{\ce{CO}} = 2p_{\ce{H2O}}$ at equilibrium.
So from substitution into Martin's equation, $\ce{p_{H_2O} + p_{CO} + p_{\ce{CO2}} + p_{\ce{H2}}}=1.5$, we find that $p_{\ce{H2O}} = p_{\ce{CO}} = \pu{0.25 bar}$.
But they are now only a third of their original concentration, and so their pressure has also decreased by a factor of 3. So their original pressure is $\pu{0.75 bar}$ each.