Call $M$ the midpoint of arc $KL$ (in the opposite half-plane of $\omega_B$ and $\omega_D$ w.r.t. $KL$). By Poncelet's Porism, there exist two points $P,Q\in\Gamma$ such that $I_B$ is the incentre of $\triangle PKL$ and $I_D$ is the incentre of $\triangle QKL$ (with $P,Q$ in the opposite half-plane of $M$ w.r.t. $KL$). By Incentre-Excentre Lemma, $I_B,I_D,K$ and $L$ all lie on the circle of center $M$ and radius $MK$.

Cute! Let the other tangent from $K$ and $L$ meet $(ABCD)$ at $K^{*}$ and $L^{*}$. By Poncelet's Porism, we have that $I_B,I_D$ are incenters of $KK^{*}L$ and $LL^{*}K$, now this is a well known lemma that $KLI_BI_D$ is cyclic.

Power of $I_B$ wrt. $\Gamma$ is $-2r_BR$, and analogously for $I_D$. Linpop finishes.

Here is a different (less clever) solution. Let $S = (AI_BI_D)\cap \Gamma \neq A$, and let $M_B$ and $M_C$ be the midpoints of arcs $BC$ and $CD$ on $\Gamma$, respectively. The key claim is that: Claim: [Harmonic Quad] We have $(S, C; M_B, M_D) = -1$. Proof. Note that $S$ is the center of the spiral similarity sending $I_BI_D\to M_BM_D$, so we have $\triangle SI_BM_B\sim \triangle SI_DM_D$. The claim follows since \[ \frac{SM_B}{SM_D} = \frac{I_BM_B}{I_DM_D} = \frac{CM_B}{CM_D}, \]where the last equality follows from Incenter / Excenter lemma. $\blacksquare$ [asy][asy] size(300); defaultpen(fontsize(10pt)); pair A, B, C, D, IB, ID, K, L, R, RR, T, U1, V1, U2, V2, K1, L1, K2, L2, MB, MD, S; A = dir(205); B = dir(125); C = dir(335); D = dir(260); IB = incenter(A, B, C); ID = incenter(A, C, D); R = extension(A, C, IB, ID); RR = rotate(90, R)*ID; T = extension(foot(IB, RR, ID), foot(ID, RR, IB), IB, ID); U1 = IP(incircle(A, B, C), CP((IB+T)/2, IB), 0); V1 = IP(incircle(A, D, C), CP((ID+T)/2, ID), 0); U2 = IP(incircle(A, B, C), CP((IB+T)/2, IB), 1); V2 = IP(incircle(A, D, C), CP((ID+T)/2, ID), 1); K1 = IP((3*U1-2*V1)--U1, circumcircle(A, B, C)); L1 = IP(V1--T, circumcircle(A, B, C)); K2 = IP((3*U2-2*V2)--U2, circumcircle(A, B, C)); L2 = IP(V2--T, circumcircle(A, B, C)); MB = circumcenter(B, IB, C); MD = circumcenter(C, ID, D); S = IP(A--T, circumcircle(A, B, C), 1); draw(A--B--C--D--cycle, orange); draw(circumcircle(A, B, C), red); draw(A--C, orange); draw(incircle(A, B, C), heavygreen); draw(incircle(A, C, D), heavygreen); draw(K2--T, lightblue); draw(circumcircle(A, IB, ID), magenta+dashed); draw(A--T, magenta+dotted); draw(A--MB--MD--cycle, heavycyan); draw(S--MB--C--MD--cycle, purple+linewidth(0.8)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$I_B$", IB, dir(90)); dot("$I_D$", ID, dir(300)); dot("$R$", R, dir(90)); dot("$T$", T, dir(270)); dot("$M_B$", MB, dir(MB)); dot("$M_D$", MD, dir(MD)); dot("$S$", S, dir(S)); dot("$K$", K2, dir(K2)); dot("$L$", L2, dir(315)); [/asy][/asy] Now we are ready to finish. Let $R$ and $T$ be the insimilicenter and exsimilicenter of $\omega_B$ and $\omega_C$, respectively. We know that $R = \overline{AC}\cap \overline{I_BI_D}$, $T = \overline{KL} \cap \overline{I_BI_D}$, and $(R, T; I_B, I_D) = -1$. Projecting the harmonic quad through $A$, we have \[ -1 = (S, C; M_B, M_C)_{\Gamma} \stackrel{A}{=} (R, \overline{AS}\cap \overline{I_BI_D}; I_B, I_D), \]which implies that $T = \overline{AS}\cap \overline{I_BI_D}$. Finally, power of a point gives \[ TI_B\cdot TI_D = TA\cdot TS = TK\cdot TL, \]so $I_B$, $I_D$, $K$, and $L$ are cyclic, as desired. Remarks: [Motivation] Unsurprisingly, this solution was found backwards. After constructing $T$, the points $K$ and $L$ become irrelevant; one just needs to show that the power of $T$ with respect to $\Gamma$ is $TI_B\cdot TI_D$. This motivates looking for circles containing $I_B$ and $I_D$, and $(AI_BI_D)$ is a natural choice.

Poncelet porism tells us that there exists $U,V$ on $(ABCD)$ such $\omega_b,\omega_c$ are the incircles of $\triangle UPQ$ and $\triangle VPQ$. Now, note \[\angle PI_BQ =\frac{\angle PUQ}{2}+90^\circ=\frac{\angle PVQ}{2}+90^\circ=\angle PI_CQ\]by Bowtie, so we are done.