It would be $\Theta \left(n\log{n} \right)$.
At each level, the array of size n is getting divided in to 2 sub arrays of size (n/4) and (3n/4) along with an extra work for choosing pivot which requires $O \left( n \right)$ time.
So recurrence will be of the form: $T(n) = T\left(\frac{n}{4}\right) + T\left(\frac{3n}{4}\right) + O\left(n\right)$
This can be solved using recursion tree.
Lower bound for this recurrence will be $\Omega \left(n\log_{4}n \right)$, & upper bound will be $O \left(n\log_{\frac{4}{3}}n \right)$,
which gives $T(n) = \Theta \left(n\log{n} \right)$.