Evaporation of a liquid (and condensation of vapour) are physical processes. We can write the equations for a liquid material $\text{M}$ as follows:
$$\ce{M_{(l)}} \rightleftharpoons \ce{M_{(g)}}$$
By the law of mass action, we can write the expression for the equilibrium constant as follows:
$$K = \dfrac{[\ce{M_{(g)}}]}{[\ce{M_{(l)}]}}$$
We typically use partial pressures for the activity of gases and unity for the activity of liquids. Thus,
$$K_{p} = \dfrac{p_{M}}{1} = p_{M}$$
Suppose we start with the liquid $\ce{M}$ in a container with an arbitrary amount of headspace. It will evaporate till the vapour exerts a pressure equivalent to $p_{M}$ (of course, $p_{M}$ depends on the temperature).
Thus, with a change in the size of a container, there will be a difference in the amount of $\ce{M}$ in the vapour phase, but not the partial pressure exerted by the vapours of $\ce{M}$ at equilibrium (a.k.a. the vapour pressure).
The shape might affect the flux at which the dynamic exchange between vapour and liquid phase $\ce{M}$ is taking place, but nothing more.