Answer
As in the discussion preceding Equation 6, if we keep K constant $K=K_{0}$, $P(L, K_{0})$ is a function of a single variable $L$, and
$\displaystyle \frac{dP}{dL}=\alpha\frac{P}{L}$ is a separable differential equation.
Integrating both sides,
$\displaystyle \int\frac{dP}{P}=\int\alpha \displaystyle \frac{dL}{L}$
$\ln|P|=\alpha\ln|L|+C$
where $C=C(K_{0})$ depends on $K_{0}$. Apply $e^{(...)}$ to both sides.
$|P|=e^{\alpha\ln|L|+C(K_{0})}\qquad$ ...both P and L are posititive, so
$P=e^{\alpha\ln L}e^{C(K_{0})}$
$=e^{\ln L^{\alpha}}\cdot e^{C(K_{0})}$
$=L^{\alpha}\cdot C_{1}(K_{0})\qquad $... another constant that depends on $K_{0}$.
This is what needed to be shown, $P(L, K_{0})$=$C_{1}(K_{0})\cdot L^{\alpha}$
Work Step by Step
All steps shown in the answer.
