In the given figure, O is the centre of a circle, BOA is its diameter and the tangent at the point

We know that a chord passing through the centre is the diameter of the circle.
∵∠BPA = 90∘    (Angle in a semi circle is 90∘)
By using alternate segment theorem
We have ∠APT = ∠ABP = 30∘
Now, In △ABP
∠PBA + ∠BPA + ∠BAP = 1800            [Angle sum property of a triangle]
⇒ 30∘ +  900 + ∠BAP = 180∘                      
⇒ ∠BAP = 60∘ 
Now, ∠BAP = ∠APT + ∠PTA
⇒ 60∘ = 30∘ +  ∠PTA
⇒ ∠PTA = 30∘
Hence, the correct answer is option (b).