Correct Answer: 70$^\circ$

Solution :
Given, an isosceles $\triangle$ ABC with AB = BC
And, $\angle$ A = 40$^\circ$ with bisectors PO and OQ of $\angle$ ABD and $\angle$ ACE.
In $\triangle$ ABC,
$\angle$A + $\angle$ B+ $\angle$ C = 180$^\circ$
Or, 40$^\circ$ + 2$\angle$C = 180$^\circ$
Or, $\angle$ B = $\angle$ C = 70$^\circ$
Now, $\angle$ ABD + $\angle$ ABC = 180$^\circ$
Or, $\angle$ ABD + 70$^\circ$ = 180$^\circ$
Or, $\angle$ ABD = 110$^\circ$
Since PO is an angle bisector, $\angle$ PBD = $\frac{110^\circ}{2}=55^\circ$
Similarly, $\angle$ QCE = 55$^\circ$
In $\triangle$ BCO,
$\angle$BOC + $\angle$ CBO+ $\angle$ BCO = 180$^\circ$
Or, $\angle$ BOC + $\angle$ PBD+ $\angle$ QCE = 180$^\circ$ (vertically opposite angles)
Or, $\angle$ BOC + 55$^\circ$ + 55$^\circ$ = 180$^\circ$
Or, $\angle$ BOC = 70$^\circ$
Hence, the correct answer is 70$^\circ$.