Power output with 46.85 kV sending, 33 kV receiving

Short line with $Z = 20.62\angle75.96^\circ = 5.00 + j20.00\ \Omega$ per phase.

Method

Take $V_R$ (line, kV) as reference. For three-phase $P + jQ$ in MW/MVAr at the receiving end:

$$V_S = V_R + \frac{Z\,(P - jQ)}{V_R}$$

Multiplying by $V_R$ and taking magnitudes:

$$|V_S|^2|V_R|^2 = (V_R^2 + RP + XQ)^2 + (XP - RQ)^2$$

Substitution

At 0.8 pf lagging, $Q = 0.75P$. With $V_R = 33$, $V_S = 46.85$, $R = 5$, $X = 20$:

$$\begin{aligned} (1089 + 5P + 15P)^2 + (20P - 3.75P)^2 &= (46.85\times33)^2 \\ (1089 + 20P)^2 + (16.25P)^2 &= 2\,390\,270 \end{aligned}$$

Expanding:

$$664.06P^2 + 43\,560P - 1\,204\,349 = 0$$

Taking the positive root:

$$P = 20.95\ \text{MW},\quad Q = 0.75P = 15.71\ \text{MVAr}$$

Check: $|33 + (5 + j20)(20.95 - j15.71)/33| = 46.85$ kV.

Sending-end power factor

$$\begin{aligned} I &= \frac{\sqrt{20.95^2 + 15.71^2}}{\sqrt3\times33} = 0.4581\ \text{kA} = 458.1\ \text{A} \\ P_{loss} &= 3I^2R = 3(458.1)^2(5) = 3.148\ \text{MW} \\ Q_{loss} &= 3I^2X = 3(458.1)^2(20) = 12.59\ \text{MVAr} \\ P_S &= 20.95 + 3.15 = 24.10\ \text{MW} \\ Q_S &= 15.71 + 12.59 = 28.30\ \text{MVAr} \\ \text{pf}_S &= \frac{24.10}{\sqrt{24.10^2 + 28.30^2}} = 0.648\ \text{lagging} \end{aligned}$$

(Check: $|S_S| = \sqrt3\times46.85\times0.4581 = 37.17$ MVA, and $\sqrt{24.10^2 + 28.30^2} = 37.17$ MVA.)

Answer: Power output ≈ 20.95 MW at 0.8 pf lag; sending-end pf ≈ 0.648 lagging.