I believe it should be $\max k=\left\lfloor\frac n2\right\rfloor$. If $A_iA_{i+1}A_{i+2}A_{i+3}$ we simply say that $P(i)$ holds. First of all, notice that if $P(i)$ holds, then $P(i+1)$ cannot hold. This is because $A_i,A_{i+4}$ must lie on the same branch of a hyperbola with foci $A_{i+1},A_{i+3}$ and on the same side of $A_{i+1}A_{i+3}$, which means that one of the points $A_i,A_{i+4}$ must lie inside the triangle formed by the other one and the two foci. This means that if $P(i)$ holds, neither $P(i+1)$ nor $P(i-1)$ can hold, so we get the upper bound $k\le\left\lfloor\frac n2\right\rfloor$. I'll just describe a construction of such a polygon with $k=\left\lfloor\frac n2\right\rfloor$ for $n=2m$ (when $n$ is odd we can modify this one, but it's a bit harder to explain). Consider a regular $m$-gon, and if, say, $A,B,C$ are consecutive vertices in this order of this $m$-gon, cut off $B$ along the segment $DE,\ D\in AB,E\in BC$ s.t. $BD=BE=x$, where $x<\frac{AB}2$ is chosen so that if $U\in AB,V\in BC$ with $AU=CV=x$, then $UDEV$ has an inscribed circle (it's clear that such an $x$ exists, by a continuity argument).

Answer is indeed $k=[\frac{n}{2}]$. Suppose that $A_1A_2A_3A_4$ and $A_2A_3A_4A_5$ have inscribed circle $\implies$ $A_1A_4+A_2A_3=A_1A_2+A_3A_4$ and $A_2A_5+A_3A_4=A_2A_3+A_4A_5$ $\implies$ $A_2A_5+A_1A_4=A_1A_2+A_4A_5$ which is not true because $A_1A_2A_4A_5$ is convex quadrilateral. Contradiction $\implies$ $k \le [\frac{n}{2}]$. As for the construction: 1. Even $n$: Construct isosceles tangential trapezium $A_1A_2A_3A_4$ with $\angle A_1A_2A_3=\angle A_2A_3A_4=\frac{n-2}{n} \pi$. Now just simply add $A_5$ such that $A_1A_2A_3A_4=A_2A_3A_4A_5$. Do this until we complete $n$-gon. Obviously $A_{i}A_{i+1}A_{i+2}A_{i+3}$ and $A_{i+2}A_{i+3}A_{i+4}A_{i+5}$ are tangential (while $A_{i+1}A_{i+2}A_{i+3}A_{i+4}$ is not) $\implies$ $k=\frac{n}{2}$. 2. Odd $n$. By same process we construct $n+1$-gon then "delete" $A_{n+1}$ and place $A_n$ such that $A_{n-2}A_{n-1}A_nA_1$ is tangential $\implies$ $k=\frac{n-1}{2}=[\frac{n}{2}]$

Solution from Twitch Solves ISL. The main claim is the following: Claim: We can't have both $A_1 A_2 A_3 A_4$ and $A_2 A_3 A_4 A_5$ be circumscribed. Proof. If not, then we have the following diagram, where $a = A_1A_2$, $b = A-2 A_3$, $c = A_3 A_4$, $d = A_4 A_5$. [asy][asy] size(9cm); pair A1 = dir(10); pair A2 = dir(40); pair A3 = dir(90); pair A4 = dir(130); pair A5 = dir(200); draw(A1--A2--A3--A4--A5, blue); label("$a$", midpoint(A1--A2), dir(A1+A2), blue); label("$b$", midpoint(A2--A3), dir(A2+A3), blue); label("$c$", midpoint(A3--A4), dir(A3+A4), blue); label("$d$", midpoint(A4--A5), dir(A4+A5), blue); draw(A1--A4, red); label(rotate(-20)*"$c+a-b$", 0.6*A4+0.4*A1, dir(A1+A4), red); draw(A2--A5, orange); label(rotate(30)*"$b+d-c$", 0.5*A5+0.5*A2, dir(A2+A5), orange); dot("$A_1$", A1, dir(A1), blue); dot("$A_2$", A2, dir(A2), blue); dot("$A_3$", A3, dir(A3), blue); dot("$A_4$", A4, dir(A4), blue); dot("$A_5$", A5, dir(A5), blue); [/asy][/asy] Then $A_1 A_4 = c+a-b$ and $A_5A_2 = b+d-c$. But now \[ A_1 A_4 + A_2 A_5 = (c+a-b) + (b+d-c) = a+d = A_1 A_2 + A_4 A_5 \]but in the picture we have an obvious violation of the triangle inequality. $\blacksquare$ This immediately gives an upper bound of $\left\lfloor n/2 \right\rfloor$. For the construction, one can construct a suitable cyclic $n$-gon by using a continuity argument (details omitted).

The answer is $k = \boxed{\lfloor \tfrac{n}{2} \rfloor}$. For an upper bound, we rely on a key claim: Claim: It is not possible for any two consecutive quadrilaterals to have an inscribed circle. Proof: Suppose FTSOC that $A_1A_2A_3A_4$ and $A_2A_3A_4A_5$ both circumscribe a circle. Moreover, let $A_1A_2 = a$, $A_2A_3 = b$, $A_3A_4 = c$, and $A_4A_5 = d$. Note that \[A_1A_4 + A_2A_5 = (a+c-b) + (b+d-c) = a+d = A_1A_2+A_4A_5,\] an obvious contradiction. $\square$ For a construction, we form cases based on the parity of $n$. If $n=2k$, we construct an equiangular polygon with alternating sides of length $1$ and $1- \cos(\tfrac{2\pi}{n})$, which works as every other quadrilateral has an inscribed circle. If $n=2k-1$, simply delete a vertex from the previous construction, preserving $k-1$ of the desired quadrilaterals.