I think that the monodromy argument they refer to is not entirely obvious and, in my opinion, is difficult to guess if you have not seen it before. I will edit later my post to give a precise reference when I can get access to my library. But let my show that their statement is true without any monodromy argument.
For any $k \geq 0$, let us denote by $S^k(X)$ the variety: $$ S^k(X) = \overline{ \{z \in \mathbb{P}^n, \ \textrm{there exists a generic $k$-uple} \ (x_1, \ldots, x_k) \in X^k, \ \textrm{such that} \ z \in \langle x_1, \ldots, x_k \rangle \} }, $$ where $\overline{A}$ is the Zariski closure of $A$. The variety $S^k(X)$ is called the variety of $k$-secants $\mathbb{P}^{k-1}$ to $X$. The variety $S^{k}(X)$ is the closure of the projection of a $\mathbb{P}^{k-1}$-fibration over an open dense subset of $X^k$, hence it is irreducible.
Now Terracini's Lemma asserts that for generic $x_1, \ldots, x_k \in X$ and for generic $z \in \langle x_1, \ldots, x_k \rangle$, we have: $$T_{S^k(X),z} = \langle T_{X,x_1}, \ldots, T_{X,x_k} \rangle.$$
Consider now $X^*$ the dual variety of $X$ (that is the closure of the set of generic tangent hyperplane to $X$). The variety $X^*$ is the closure of the image in $\left(\mathbb{P}^n\right)^*$ of the projectivization of the normal bundle of $X$ over the dense open subset $X_{smooth}$. Hence it is irréducible. We also define $X^*_k$ to be: $$ X^*_k = \overline{ \{ H \in X^*,\ \textrm{there exists a generic} \ (x_1,\ldots, x_k) \in X^k, \ \textrm{such that} \ T_{X,x_1} \subset H, \ldots, T_{X,x_k} \subset H \}}$$
Terracini's lemma implies that $X^*_k = S^k(X)^*$, where $S^k(X)^*$ is the projective dual to $S^k(X)$. We let $$p : \mathbb{P}^n \times (\mathbb{P}^n)^* \longrightarrow \mathbb{P}^n$$ and $$ q : \mathbb{P}^n \times (\mathbb{P}^n)^* \longrightarrow \left(\mathbb{P}^n \right)^*$$ the natural projections and set $I_{X^*_k} = q^{-1}(X^*_k)$. The generic contact locus they claim to be equidimentional is the generic fiber of $q$. By hypothesis, we have: $$ p(I_{X_k^*}) = X.$$
Furthermore, Terracini's lemma implies that the generic fiber (over $x \in X$) of $p|_{I_{X^*_k}}$ is $S(T_{X,x}, S^{k-1}(X))^*$, which is irreducible, as the projective dual of an irreducible variety. As the consequence, there is only one irreducible component of $I_{X^*_k}$ which dominates $X$. We denote this component by $Z$. Note also that $Z$ is the unique irreducible component of $I_{X^*_k}$ which dominates $X^*_k$. Indeed, by hypothesis, if $H \in X^*_k$ is generic then $p(q^{-1}(H))$ contains a generic point of $X$ (and we have seen that $Z$ is the only irreducible components of $I_{X^*_k}$ which dominates $X$).
The Theorem of the generic fiber guarantees that the generic fiber of the restriction $q|_{Z} : Z \longrightarrow X^*_k$ is equidimensional and so is the geenric fiber of $q$ since $Z$ is the only irreducible component of $q^{-1}(X^*_k)$ which dominates $X^*_k$.