Answer
Hence, $P_1$ is negative and $P_2$ is positive.
Work Step by Step
The Divergence Theorem states that $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV $
Here, $S$ is a closed surface and $E$ is the region inside that surface.
$div F=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}+\dfrac{\partial c}{\partial z}$
When the net flow of water is inwards, then the divergence at that point will be negative. However, when the net flow of water is outwards, then the divergence at that point will be positive.
Also, when there is no net flow of water inwards or outwards, then the divergence at that point will be zero.
We see at the point $P_1$ that the net flow of water is inwards. Thus, the divergence at that point $P_1$ will be negative. At the point $P_2$, the net flow of water is outwards, so the divergence at that point will be positive.
Hence, $P_1$ is negative and $P_2$ is positive.
