Hit for the cycle

Let $\angle ABC = \theta.$ Then since $ABCD$ is cyclic we have that $\angle CDA = 180^{\circ} - \theta.$ Also, since $AD$ is a diameter we have that $\angle ACD = 90^{\circ}.$

Using the Cosine rule on $\Delta ABC$ we see that $|AC|^{2} = 2a^{2}(1 - \cos(\theta)).$

Next, as $\Delta ACD$ is right-angled at $C$ we have that $|AC|^{2} = d^{2} - b^{2}$ and that

$\cos(\angle CDA) = \dfrac{b}{d} \Longrightarrow \cos(\theta) = -\dfrac{b}{d}.$

Combining these results, we see that

$d^{2} - b^{2} = 2a^{2}(1 + \dfrac{b}{d}) \Longrightarrow d(d^{2} - b^{2}) = 2a^{2}(d + b) \Longrightarrow 2a^{2} = d(d - b),$ (i).

Now clearly $d \gt b,$ so we cannot have $d = 1.$ So look at $d = 2.$ Plug this into (i) to get $2a^{2} = 2(2 - b) \Longrightarrow b + a^{2} = 2.$ Since $a,b$ must be positive integers we would require that $a = b = 1,$ but as it was specified that $a \ne b$ we cannot have $d = 2.$

Now look at the case where $d \ge 3$ is prime. Looking at equation (i), we see that since $d \ge 3$ cannot divide $2$ it must divide $a^{2}.$ But since $d$ is prime it must then divide $a$ . But since it is clear that $d \gt a$ we cannot have that $d|a,$ and thus $d$ cannot be a prime.

So now look at the cases where $d$ is a composite number, starting with $4, 6, 8, 9, .....$

With $d = 4$ equation (i) becomes $a^{2} = 2(4 - b).$ For $b = 1$ or $b = 3$ we have, respectively, $a^{2} = 6$ or $a^{2} = 2,$ making $a$ non-integral. With $b = 2$ we end up with $a = 2,$ violating the condition $a \ne b.$

With $d = 6$ equation (i) becomes $a^{2} = 3(6 - b).$ Trying values for $b$ of $1,2,4,5$ yield non-integral values for $a,$ and trying $b = 3$ yields $a = 3.$ Thus $d$ cannot be $6.$

With $d = 8$ equation (i) becomes $a^{2} = 4(8 - b).$ Trying $b = 7$ gives $a = 2,$ thus giving us a valid solution with $ABCD$ having a perimeter of $2 + 2 + 7 + 8 = 19.$ No other value of $b$ yields an integral value of $a.$

Now with $d = 9$ we can go through the cases for each value for $b$ to find that $b = 1, a = 6$ and $b = 7, a = 3$ are valid solutions. But these solutions each yield a perimeter of $22,$ , so $19$ is still our minimum.

With $d \gt 9$ and composite, clearly $(2a + b) \gt 9$ as well, forcing any possible perimeters we find for such values for $d$ to exceed $20.$ Thus we can conclude that the minimum possible perimeter is $\boxed{19}.$

$\angle AOB = \angle BOC = \angle ODC$ because central and inscribed angles.

$\angle OCD = \angle ODC$ because $\triangle COD$ is isosceles. So, $BO$ is parallel to $CD$ . $O$ is midpoint of $AD$ , so $OP = \frac{b}{2}$ (please refer to draw above)

$BO = AO = DO = \frac{d}{2}$ . Then, $BP = BO - PO = \frac{d-b}{2}$

$\triangle BAP$ is right because ABCO is a kite and its diagonals are perpendicular. It gives us $sin(\frac{\alpha}{2}) = \frac{d-b}{2a}$

$\triangle ACD$ is right too and gives us $cos(\alpha) = \frac{b}{d}$

Now is time to remember this trigonometric formulae for double angles: $cos(\alpha) = 1 - 2sin^{2}(\frac{\alpha}{2})$

Using it, we have: $\frac{b}{d} =1 - 2 ( \frac{d-b}{2a} )^{2}$ . Doing easy math, we get $d(d-b) = 2a^{2}$ .

As a, b, d are integer and (a + a + b + d) the minimum possible, I started with a = 1, then $d(d-b) = 2 \rightarrow d = 2$ and $b = 1$ but we cannot have a = b.

Next try was a = 2 and then $d(d-b) = 8 \rightarrow d = 8$ and $b = 7$ .

Others tries show me larger perimeters, so the minimum one is 2 + 2 + 7 + 8 = $\boxed {19}$ .

Congratulations for the beautiful problem. Was very nice to solve it.