Answer
$P(L, K_{0})$=$C_{1}(K_{0})\cdot L^{\alpha}$
Work Step by Step
We need to keep $K$ as constant, $K=K_{0}$,
Here, $P(L, K_{0})$ shows a function of a single variable $L$, and
$ \dfrac{dP}{dL}=\alpha\frac{P}{L}$ shows a separable differential equation.
Further,
$ \int\dfrac{dP}{P}=\int\alpha \dfrac{dL}{L}$
This gives: $\ln|P|=\alpha\ln|L|+C$
Here, $C=C(K_{0})$ depends on $K_{0}$.
$|P|=e^{\alpha\ln|L|+C(K_{0})}$
or, $P=e^{\alpha\ln L} \times e^{C(K_{0})}$
or $=e^{\ln L^{\alpha}}\times e^{C(K_{0})}$
or, $|P|=L^{\alpha}\cdot C_{1}(K_{0});$ Another constant that depends on $K_{0}$.
Hence, we have $P(L, K_{0})$=$C_{1}(K_{0})\cdot L^{\alpha}$
