Denote the radius of circle K $r_1$, the radius of circle L $r_2$ then $r_2(\sec \theta+1)=c \sin\theta $ , similarily $r_1(\csc \theta+1)=c \cos \theta $ and $p<=1+\frac{2}{\sin\theta+\cos\theta}$ Since $1+\frac{2}{\sin\theta+\cos\theta}>=1+\sqrt2$ , we have $p<=1+\sqrt 2$

Alternative without trig: First consider circle $k$ and call the center $K$. Notice that it must be tangent to the side $b$ at $C$ (because $C$ is $90^{\circ}$. Let the circle also be tangent to side $c$ at $C'$. Thus we have that $KC = r$ and $KC' = r$ We have that $\angle{CAB} = \angle{CAB}, \angle{AC'K} = \angle{ACB} = 90^{\circ}$ thus the triangles are similar by $AA(A)$ similarity. This means that $\frac{KC'}{BC} = \frac{AK}{AB}$ We have that $AK = AC - KC = b-r$. When we replace with variables we get $\frac{r}{a} = \frac{b-r}{c} \rightarrow cr = ab - ar \rightarrow r = \frac{ab}{a+c}$ Next, we do the same with circle $L$. Through the exact same process we get that $t = \frac{ab}{b+c}$ By replacing it into said inequality, we get that $\frac{b+c}{ab} + \frac{a+c}{ab} \ge p(\frac{a+b}{ab}) \rightarrow b+c + a+c \ge p(a+b) \rightarrow 2c \ge (p-1)(a+b)$ Let $k=p-1$. Thus we have $2\sqrt{a^2+b^2} \ge k(a+b)$ by squaring both sides (since we know that all the values are positive and we are assuming the greatest $p$ is greater than 1) we have $4a^2+4b^2 \ge k^2a^2 + 2k^2ab + k^2b^2 \rightarrow (4-k^2)a^2 - 2k^2ab + (4-k^2)b^2 \ge 0$. Assume $k^2 > 4-k^2$. That means $k^2 = 4-k^2+c$ for some positive number $c$. Thus the equation becomes $(4-k^2)(a^2-2ab+b^2) - c \ge 0 \rightarrow (4-k^2)(a-b)^2 - c \ge 0$ which is obviously false if $a=b$ This means that $k^2 \le 4-k^2 \rightarrow 2k^2 \le 4 \rightarrow k^2 \le 2 \rightarrow k \le \sqrt{2}$ We plug this in to see if the inequality is satisfied and in fact if $k = \sqrt{2}$ the inequality becomes $2a^2 - 4ab + 2b^2 = 2(a-b)^2 \ge 0$ which is true by the trivial inequality. Since $k$ is maximized at $\sqrt{2}, k+1 = p$ is maximized at $\boxed{\sqrt{2}+1}$