Answer
$L \dfrac{\partial P}{\partial L}+K \dfrac{\partial P}{\partial K} =(\alpha+\beta)P$
Work Step by Step
$ \dfrac{\partial P}{\partial L}=\alpha bL^{\alpha-1} \times K^{\beta}$
and $ \dfrac{\partial P}{\partial K}=\beta bL^{\alpha} \times K^{\beta-1}$
Consider the LHS such as: $(L) \dfrac{\partial P}{\partial L}+(K) \dfrac{\partial P}{\partial K}= L \times (\alpha bL^{\alpha-1}K^{\beta})+K \times (\beta bL^{\alpha}K^{\beta-1})=\alpha b L^{\alpha}K^{\beta}+\beta b L^{\alpha}K^{\beta}=(\alpha+\beta)b L^{\alpha} K^{\beta}$
This implies that $L \dfrac{\partial P}{\partial L}+K \dfrac{\partial P}{\partial K} =(\alpha+\beta)P$
