From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top…

Correct Figure.
In $\triangle BPA$
$\tan 45^\circ = 1 = \frac{20}{x}$
$\Rightarrow x = 20 \text{ m} \dots (i)$
Now, In $\triangle CPA$
$\tan 60^\circ = \sqrt{3} = \frac{h+20}{x}$
$\Rightarrow h + 20 = x\sqrt{3} \dots (ii)$
Solving (i) and (ii)
$h = 20 (\sqrt{3} - 1) \text{ m}$
$\therefore$ Height of the tower is $20 (\sqrt{3} - 1) \text{ m}$.