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?

Actually the background is the probability of default.