Since there is no battery in the circuit I must assume that some sort of chemical reaction is moving positive charges from $B$ to $A$.
The cases to consider are that the electrodes $AP$ and $BQ$ have no resistance or the electrodes have some resistance.
If the electrodes have no resistance then $V(P) = V(A)$ and $V(Q) = V(B)$
So the potential difference across the electrodes is zero.
If the electrodes have the same resistance and given the direction of flow of the positive charges then the potential of $A$ must be greater than the potential of $P$ and the potential of $Q$ must be greater than the potential of $B$.
This assumes that no chemical reaction is going on inside the electrodes.
So $V(P)-V(A)$ is less than zero and $V(Q)-V(B)$ is greater than zero because positive charges not inside a cell flow from from a region of high potential to a region of low potential.
This is the opposite to your statement.
If you assign a value of $V$ to $V(P)-V(A)$ then $V$ will be a negative value.
Then as the resistances of the two electrodes are the same $V(Q)-V(B) = -V$ and as $V$ is the numerical value of $V$ is negative then $-V$ will be positive.
To sum up:
$V(P)-V(A)$ is negative ($P$ is at a lower potential relative to $A$) and $V(Q)-V(B)$ is positive ($Q$ is at a higher potential relative to $B$).