suppose $$dA = \mu Adt + \sigma AdX.$$ is a geometric Brownian motion. One says that the Probability $P(A,t)$ of $A$ reashing the critical level $K(t)$ before maturity: $$\dfrac{\partial P}{\partial t} + \dfrac{1}{2}\sigma^2A^2\dfrac{\partial^2 P}{\partial A^2}+\mu \dfrac{\partial P}{\partial A} = 0$$ I know this is the Kolomogorov backward equation for transition density, but why this is true for probability function here? It seems we should replace CDF by PDF i.e replace $P$ by $\dfrac{\partial P}{\partial x}$?

Actually the background is the probability of default, Paul Wilmott on Quantitative Finance Volumne II Page641