Find the transition probability function P(y,t,x,s) for Brownian motion with drift B(t)+t.
I have already know the standard Brownian motion transition fuction is N（0，t），whose drift term is constant。
but i can't see how to transform the drift（B(t)+t）to be a constant.
You can write this using the usual transition density that you know: \begin{align} \mathbb{P}(B_t+t\leq y | B_s+s=x) &= \mathbb{P}(B_t\leq y-t | B_s=x-s) \\ &=\mathbb{P}(B_t-B_s\leq y-t-(x-s) | B_s=x-s)\\ &=\mathbb{P}(B_{t-s}\leq y-t-x+s), \end{align} where the last equality follows by the independent increments property of Brownian motion, or the Markov property. You have already said that Brownian motion is Gaussian, so this last line is equal to the probability that a $N(0,t-s)$ random variable is less than or equal to $y-t-x+s$. We know that this is equal to $\Phi\left(\frac{y-t-x+s}{\sqrt{t-s}}\right)$, where $\Phi$ is the cdf of a $N(0,1)$ random variable.
To find the density we differentiate. Denote the density by $p(y,t;x,s)$ as you have, and let $\phi$ be the pdf of the $N(0,1)$ dsitribution. Then, \begin{align} p(y,t;x,s)&=\frac{\mathrm{d}}{\mathrm{d}y}\Phi\left(\frac{y-t-x+s}{\sqrt{t-s}}\right) \\ &=\frac{1}{\sqrt{t-s}}\phi\left(\frac{y-t-x+s}{\sqrt{t-s}}\right) \\ &=\frac{1}{\sqrt{2\pi(t-s)}}\exp\left(-\frac{(y-x-(t-s))^2}{2(t-s)}\right). \end{align}
In general, the density of $\sigma^2 B_t +\mu t$ is $\frac{1}{\sqrt{2\pi \sigma^2 t}}\exp\left(-\frac{(x-\mu t)^2}{2\sigma^2 t}\right)$.