I think I may have reason for your reviewer's caution. You are updating the proportional gain of a PID controller, and you are concerned with the stability of your update law. Restating your update: $$K_p(t+1)=K_p+e(t)(\mu_1(Pe(t))+\mu_4(Pe(t))).$$ You have re-stated that $e(t)$ is the difference, which means that the error is not positive semi-definite ($e(t)\geq 0$). However, the membership functions $\mu_1$ and $\mu_4$ are positive semi-definite; this implies that $E(t)=e(t)(\mu_1(Pe(t))+\mu_4(Pe(t)))$ is either positive or negative semi-definite, which means that $K_p$ grows unbounded in some region $D$.
At this point, it is clear that the update law is not stable in some region (in the sense of Lyapunov) for $e(t)\neq 0$. We can substantiate this with discrete-time Lyapunov analysis.
$Proof:$ Dropping the time dependence for clarity, let $$V(K_p)=\frac{1}{2}K_p^2$$ be a candidate Lyapunov function. The rate of change along solutions is given by $$\begin{align}\nabla V(K_p)&=V(K_p+E)-V(K_p)\\&=\frac{1}{2}(K_p+E)^2-\frac{1}{2}K_p^2\\&=K_pE+\frac{1}{2}E^2\end{align}.$$
For stability of the system, we must have $\nabla V(K_p)<0$. This implies $$K_pE+\frac{1}{2}E^2<0$$$$\to K_p<-\frac{1}{2}E.$$ We can conclude from converse Lyapunov results that the system is unstable for at least $K_p>-\frac{1}{2}E$, which is an obvious reason for concern. There may be a better Lyapunov candidate function to demonstrate this result, but we can be sure that the system is locally unstable.
