Answer
$\bf Increases.$
Work Step by Step
First of all, we need to draw the force diagram of the ball in both cases, when it has a charge of $+q$ and when it has no charge at all.
We assume that the friction force between the wire and the pivot point is negligible in both cases.
$\textbf {Note the new coordinates}$ at this moment in the two figures below.
In the first case, when the ball is charged, the net force exerted on the ball in the $y$-dirction is zero while the net force exerted on it in the $x$-direction is not zero.
$$\sum F_y=F_T-mg\cos\theta-F_E\cos\theta=0$$
where $F_E=qE$, where $q$ is the charge's amount in the ball, and $F_T$ is the tension force.
$$ F_T-mg\cos\theta-qE\cos\theta=0\tag 1$$
And
$$\sum F_x=mg\sin\theta+qE\sin\theta=ma_r$$
Hence,
$$a_r= \left[g+\dfrac{qE }{m}\right]\sin\theta\tag 2$$
In simple pendulum, we know that $a=g\sin\theta$ and that $T=2\pi \sqrt{\dfrac{L}{g}}$, so in this case,
$$\boxed{T=2\pi \sqrt{\dfrac{L}{g+\frac{qE }{m}}}}$$
As $q$ approaches zero, $T$ increases.
And when $q=0$, $T$ reaches its maximum value of $2\pi\sqrt{L/g}$, as in the second figure below.
Therefore, the period increases when the ball is discharging.
