Correct Answer: 63º

Solution : A circle with the centre O and diameter AB.
P and Q are two points on the circle, and $\angle$POQ = 54$^\circ$
Concept used:
The angle subtended by the chord at the circumference would be half of the angle subtended at the centre
Calculation:
We know that the angle subtended by the diameter on the circumference is 90
⇒ $\angle$APB = 90$^\circ$
The angle subtended by chord PQ at the circumference would be half of the angle subtended at the centre
$\angle$QAP = $\frac{54^\circ}{2}$
⇒ $\angle$QAP = 27$^\circ$
In the Δ APC
$\angle$APC = $\angle$APB = 90$^\circ$
$\angle$CAP = $\angle$QAP = 27$^\circ$
We know that the sum of all angles in a triangle is 180$^\circ$
$\angle$PCA + $\angle$APC + $\angle$CAP = 180$^\circ$
$\angle$PCA + 90$^\circ$+ 27$^\circ$ = 180$^\circ$
⇒  $\angle$PCA = 180$^\circ$ – 117$^\circ$
⇒  $\angle$PCA = 63$^\circ$
$\therefore$ The required value of $\angle$PCA is 63$^\circ$
Hence the correct answer is 63$^\circ$