Suppose $AC=q$, let us solve for all the sides in terms of $q$:
$$\frac{\sin(135^\circ)}{q}=\frac{\sin(30^\circ)}{AD}\implies AD=\frac{q}{\sqrt2}$$
$$\frac{\sin(15^\circ)}{CD}=\frac{\sin(135^\circ)}{q}\implies\,CD=\frac{q(\sqrt3-1)}{2}$$
Since $CD=BD$, then $BC=q(\sqrt3-1)$. Then we need to find $AB$:
$$AB^2=q^2+\left(q\left(\sqrt3-1\right)\right)^2-2q^2(\sqrt3-1)\cos(30^\circ)\implies AB=q\sqrt{2-\sqrt{3}}$$
Then we just need to solve for $x$ in:
$$\frac{\sin x}{\frac{q(\sqrt3-1)}2}=\frac{\sin 45^\circ}{q\sqrt{2-\sqrt3}}\implies \sin x=\frac{\sqrt3-1}{2\sqrt{4-2\sqrt3}}$$
The nested radical in the bottom looks ugly, but remember that the radical $\sqrt{a\pm b\sqrt{c}}=\sqrt d\pm\sqrt e$ if and only if $\sqrt{a^2- b^2c}\,\,$ is rational.
Therefore, we have:
$$\sin x=\frac{\cancel{\sqrt3-1}}{2\cancel{(\sqrt3-1)}}=\frac12 \implies$$
$$\bbox[10px, border: 2px black solid]{\therefore x=\arcsin{\frac12}=\frac\pi6=30^\circ}$$
Thale's theorem states $\angle ACB$ is a right angle since $AB$ is a diameter. As $\angle BAT$ is also a right angle due to $PAT$ being tangent to the circle at $A$ (as stated in Tangent lines to circles, "The radius of a circle is perpendicular to the tangent line through its endpoint on the circle's circumference."), you have $\angle CAB = 90^{\circ} - x$ (as you already stated in your question) and, thus, $\angle ABC = x$ due to the sum of the angles in a triangle being $180^{\circ}$.
Best Answer
Hint: $\angle ACB$ is 90 degrees because AB is the diameter of the circle.