The enthalpy difference between the two phases at any temperature $$T$$ can be obtained by integrating the heat-capacity difference from 0 K to $$T$$.

For the given data, the heat-capacity difference is constant:
$$C_{P,\beta}-C_{P,\alpha}=1\ \text{J mol}^{-1}\text{K}^{-1}$$

Starting from 0 K (where both phases may be taken to have the same enthalpy), the enthalpy difference at temperature $$T$$ is

$$\begin{aligned} H_\beta-H_\alpha &= \int_{0}^{T}\bigl(C_{P,\beta}-C_{P,\alpha}\bigr)\,dT \\ &= (C_{P,\beta}-C_{P,\alpha})\int_{0}^{T} dT \\ &= (C_{P,\beta}-C_{P,\alpha})\,T \end{aligned}$$

Substituting the numerical values for the required temperature $$T=300\ \text{K}$$:

$$H_\beta-H_\alpha = 1\ \text{J mol}^{-1}\text{K}^{-1}\times 300\ \text{K} = 300\ \text{J mol}^{-1}$$

Hence, the enthalpy change at 300 K is 300 J mol$$^{-1}$$.