Auto-ionisation and Dilution
Recall that $\mathrm p\ce{OH} = -\log_{10}\space [\ce{OH-}]$
Where $\ce{[OH-]}$ is the concentration of hydroxide ions.

Suppose a solution has $10^{-7}$ moles of $\ce{[OH-]}$ ions. Then this implies $\mathrm p\ce{OH} = -\log_{10}\space [\ce{OH-}]$ and $\mathrm p\ce{OH} = -\log_{10}\space [10^{-7}]$ which is equal to $7$.

Suppose a solution has $10^{-4}$ moles of $\ce{[OH-]}$ ions. Then this implies $\mathrm p\ce{OH} = -\log_{10}\space [\ce{OH-}]$ and $\mathrm p\ce{OH} = -\log_{10}\space [10^{-4}]$ which is equal to $4$.


So as you can see, lower the $\mathrm p \ce{OH}$, the higher the concentration of $\ce{OH-}$ ions. Le Chatelier's principle is not violated.
(How I like to remember this is - the strongest acids have the lowest $\mathrm pK_\mathrm a$. So that extends to $\mathrm p\ce{OH}$ too. This is all due to the fact that $\mathrm p\ce{OH}$ is the negative logarithm of the $\ce{[OH-]}$ concentration in solution.)