Here is a derivation for the Ornstein-Uhlenbeck process. Solution to the SDE $$dX_t = \theta(\mu-X_t) dt + \sigma dW_t$$ subject to the initial condition $X_0=x$ has the form $$X_t= \mu + (x - \mu)e^{-\theta t} + \sigma\int_0^t e^{-\theta (t-s)}dW_s.\qquad$$ We need to calculate density $p(t,x,y)$ of the conditional distribution $(X_t|X_0=x)$.

$X_t$ is normally distributed for each $t>0$. The conditional expectation is $$E[X_t|X_0=x]=\mu + (x - \mu)e^{-\theta t}.$$ The conditional variance is $$Var[X_t|X_0=x] = \sigma^2E\left[(\int_0^t e^{-\theta (t-s)}dW_s)^2 \right] = \sigma^2 E\left[\int_0^t e^{-2\theta (t-s)}ds \right]$$ $$=\frac{\sigma^2}{2\theta}(1-e^{-2\theta t}).$$ Hence, we have that $$(X_t|X_0=x)\sim N\left(\mu + (x - \mu)e^{-\theta t},\frac{\sigma^2}{2\theta}(1-e^{-2\theta t})\right),$$ and, finally, $$p(t,x,y)=\frac{1}{\sqrt{\pi\sigma^2(1-e^{-2\theta t})/\theta}}\exp \left[-\frac{(-y-\mu-(x - \mu)e^{-\theta t})^2}{\sigma^2(1-e^{-2\theta t})/\theta}\right].$$

Here is a derivation for the Ornstein-Uhlenbeck process. Solution to the SDE $$dX_t = \theta(\mu-X_t) dt + \sigma dW_t$$ subject to the initial condition $X_0=x$ has the form $$X_t= \mu + (x - \mu)e^{-\theta t} + \sigma\int_0^t e^{-\theta (t-s)}dW_s.\qquad$$ We need to calculate density function $p(t,x,y)$ of the conditional distribution $(X_t|X_0=x)$.

$X_t$ is normally distributed for each $t>0$. The conditional expectation is $$E[X_t|X_0=x]=\mu + (x - \mu)e^{-\theta t}.$$ The conditional variance is $$Var[X_t|X_0=x] = \sigma^2E\left[(\int_0^t e^{-\theta (t-s)}dW_s)^2 \right] = \sigma^2 E\left[\int_0^t e^{-2\theta (t-s)}ds \right]$$ $$=\frac{\sigma^2}{2\theta}(1-e^{-2\theta t}).$$ Hence, we have that $$(X_t|X_0=x)\sim N\left(\mu + (x - \mu)e^{-\theta t},\frac{\sigma^2}{2\theta}(1-e^{-2\theta t})\right),$$ and, finally, $$p(t,x,y)=\frac{1}{\sqrt{\pi\sigma^2(1-e^{-2\theta t})/\theta}}\exp \left[-\frac{(-y-\mu-(x - \mu)e^{-\theta t})^2}{\sigma^2(1-e^{-2\theta t})/\theta}\right].$$