Trapezium angle given, relation between sides 
How do i check other options? Should i apply triangular inequality ? Also is there some way that all options are checked?

Assume that (D) is true. Then, since you have established that (C) is true, we must have $BC=AC \implies \theta =60°$ which is clearly false.
If (A) was true, then $AD=CD\implies \angle ACD =60° \implies \angle ADC =60° \implies \angle PDC =120° \implies PCD =0°$ which again, is false.
If (B) was true, then $BC=CD \implies \angle BCD=\theta$. 
Now, in $\triangle ABO$, ($O$ being the intersection of the diagonals of the quad.)
$\angle ABL = 60°-\theta \implies \angle AOB =120° \implies \angle AOD =60°\implies \angle ADO=60°\implies \angle PDC =120°-\theta \implies \angle PCD =\theta$
Now consider $\triangle BCD$. By the exterior angle property we have that $2\theta =\theta \implies \theta=0°$ 
False.
