Given that,

CP ‖ BQ

Produce CP to E

So, PE ‖ BQ and AB cuts them

∠QBE = ∠CBA = 105^o (Corresponding angles)

In

∠CEA + ∠ECA + ∠EAC = 180^o

105^o + ∠ECA + 25^o = 180^o

∠ECA = 50^o

∠PCA + ∠ECA = 180^o (Linear pair)

x + 50^o = 180^o

x = 130^o