a) Considering the figure shown above, the remaining cases are proved similarly. Let $T$ be the midpoint of $NP$. We have $(I)$ as the circle $A$-mixtilinear tangent to triangle $AA_bA_c$, so according to Sawayama's lemma, $T$ is the incenter of triangle $AA_bA_c$. By angle chasing, we conclude that
$$\triangle TA_bP \sim \triangle A_cTN.$$
It follows that
$$A_bP \cdot A_cN = TP \cdot TN = \frac{NP^2}{4}.$$
By AM-GM inequality, one can get
$$A_bP + A_cN \ge 2\sqrt{A_bP \cdot A_cN} = NP.$$
Similarly,
$$PB_a + MB_c \ge MP, \quad NC_a + MC_b \ge MN$$
From these, we conclude the required inequality.

b) Let $X$ be the intersection of $B_aB_c$ with $C_aC_b$, $Y$ be the intersection of $C_aC_b$ with $A_bA_c$ and $Z$ the intersection of $A_bA_c$ with $B_aB_c$. Let $\triangle$ be the projection triangle of $H$ corresponding to triangle $ABC$. Then $H$ is the incenter of $\triangle$ and the midpoint $OH$ is the circumcenter of $\triangle$. In this solution, we will show that $I$ is the center of the circle inscribed in the triangle $XYZ$ and that the two triangles $XYZ$ and $\triangle$ have corresponding parallel sides. From there, $IK \parallel OH$.

First, to prove that triangles $XYZ$ and $\triangle$ have corresponding sides parallel, we show that $A_bA_cCB$ is a cyclic quadrilateral, and then $A_bA_c$ is anti-parallel to $BC$ in $\angle BAC$ so it is parallel to the line connecting the foot of the vertex $B$ and $C$. Indeed, considering the inversion of the center $A$, power $AP^2$ which is denoted by $\mathcal{I}_A$. This inversion preserves $(I)$ and maps $A' \mapsto D$.

so the image of $(AA_bA_cA')$ passes through $D$ and touches $(I)$ so
$$\mathcal{I}_A : (AA_bA_c) \mapsto BC.$$
Therefore, the image of $A_b$ lies on $AB$ and $BC$, so $B$ is the image of $A_b$ through $\mathcal{I}_A$ which infer
$$\overline{AA_b} \cdot \overline{AB} = AP^2.$$
Similarly, $\overline{AA_c} \cdot \overline{AC} = AN^2 = AP^2$ so $A_bA_cCB$ is a cyclic quadrilateral. So two triangles $XYZ$ and $\triangle$ have corresponding sides are parallel.

Finally, we need to prove that $I$ is the incenter of triangle $XYZ$ or equivalently $XI$ is the angle bisector of $\angle B_cXC_b$. On the other hand, similar to the above proof, then $B_aB_cCA$ and $C_aC_bBA$ are cyclic quadrilaterals, so
$$\angle XC_bB_c = \angle BAC = \angle XB_cC_b$$
leads to $XC_b = XB_c$. So it suffices to show that $M$ is the midpoint of $B_cC_b$ and $X, M, I$ lie on the bisector $\angle YXZ$. By considering the inversions of centers $B$ and $C$ preserving $(I)$ similar to the above, we can show that
$$BB_c \cdot BC = BM^2, \quad CC_b \cdot CB = CM^2$$
and it leads to
$$MB_c = MB - BB_c = MB - \frac{MB^2}{BC} = \frac{MB \cdot MC}{BC}.$$
The length of $MC_b$ can be calculated in the same way and then $MB_c = MC_b$. Thus $XM$ is perpendicular bisector of $B_cC_b$ so $XM \perp BC$ and $I$ lies on $XM$. So $XI$ is the angle bisector of $\angle YXZ$. Similar to the vertices $Y, Z$, we can conclude that $I$ is the incenter of the triangle $XYZ$. This finishes the proof. $\square$