Answer
$L \dfrac{\partial P}{\partial L}+K \dfrac{\partial P}{\partial K} =(\alpha+\beta)P$
Work Step by Step
Here, $L$ is the variable and all other variables can be treated as constants.
$ \dfrac{\partial P}{\partial L}=\alpha bL^{\alpha-1}K^{\beta}$ ;
and $ \dfrac{\partial P}{\partial K}=\beta bL^{\alpha}K^{\beta-1}$
Consider the LHS as: $L \times \dfrac{\partial P}{\partial L}+K \times \dfrac{\partial P}{\partial K}= L(\alpha bL^{\alpha-1}K^{\beta})+K(\beta bL^{\alpha}K^{\beta-1})$
$=\alpha b \times L^{\alpha}K^{\beta}+\beta b \times L^{\alpha}K^{\beta}$
$=(\alpha+\beta)b \times L^{\alpha} \times K^{\beta}$
$=(\alpha+\beta)P$
Hence, $L \dfrac{\partial P}{\partial L}+K \dfrac{\partial P}{\partial K} =(\alpha+\beta)P$
