Calculating Active power in 3 Phase
You are using \$I_{LINE}\$ in both.

There is no dot, so the current that flows in phase must flow in transmission line.  So in a star or wye, \$I_{LINE}\ =\ I_{PHASE}\$ (and \$V_{LINE}\ =\  √ (3)\ V_{PHASE}\$).  You show this in your answer.

In a delta, \$I_{LINE}\ =\ √ (3)\ I_{PHASE}\$ (and \$V_{LINE}\ =\  V_{PHASE}\$).  A component of two phase currents make up the line current.  There is a dot.

$$P_T = √ (3)\ V_{LINE}\  I_{LINE}\ cos\ θ $$
$$P_T = 3\ V_{PHASE}\  I_{PHASE}\ cos\ θ $$

So in your first answer, \$V_{L-L}\$ which is line voltage \$V_{LINE}\ =\ 415V\$, which means a \$V_{PHASE}\ =\ 415V / √ (3)\ =\ 239.6V\$.
$$ I_{PHASE}\ =\ V_{PHASE} / Z\ = 239.6V / 10Ω = 24.0A$$
$$ P\ = I_{PHASE}^2\ R  = (24.0A)^2\ x\ 8Ω\ =\ 4.59kW$$
$$ P_T\ = 3\ P  = 3\ x\ 4.59kW\ =\ 13.8kW$$