We know that $$\angle ACB=\angle PCB$$
In $$\triangle PCB$$
Using the angle sum property
$$\angle PCB+\angle BPC+\angle PBC=180^o$$
We know that $$\angle APB$$ and $$\angle BPC$$ are linear pair
By substituting the values
$$\angle PCB +(180^o -110^o)+25^o =180^o$$
On further calculation
$$\angle PBC +70^o+25^o =180^o$$
$$\angle PCB +95^o =180^o$$
By subtraction
$$\angle PCB =180^o -95^o$$
So we get
$$\angle PCB =85^o$$
We know that the angles in the same segment of a circle equal
$$\angle ADB=\angle ACB =85^o$$
Therefore, the value of $$\angle ADB$$ is $$85^o$$