Define the load voltage phasor

Since the load is resistive, the load voltage $V_L$ is in phase with the current $I$. We take $V_L$ as the reference phasor: $V_L = 300 + j0$ V.

$V_L = 300 \angle 0°\text{ V}$

Calculate the sending end voltage

The sending end voltage $V_S$ is the sum of the load voltage and the voltage drop across the impedance: $V_S = V_L + I \times Z_{line}$. Here, $I = 500 \angle 0°$ A and $Z_{line} = j0.8$ Ω.

$V_S = 300 + (500 \angle 0°)(j0.8) = 300 + j400$

Calculate the sending end power factor

The power factor angle $\phi_S$ is the angle of the sending end voltage $V_S$. Since $V_S = 300 + j400$, the phase angle is $\tan^{-1}(\frac{400}{300}) = \tan^{-1}(1.33) = 53.13°$. The power factor is $\cos(53.13°) = 0.6$. Since the imaginary component is positive, the current lags the voltage.

$\cos(\phi_S) = \cos(\tan^{-1}(\frac{400}{300})) = 0.6 \text{ lagging}$