So say I have the tensor of type ${1}\choose{1}$, with $T \in T_p(\mathbb{R}^n) \otimes T_p^*(\mathbb{R}^n)$ where $$T = T^{a}_{b} \frac{\partial}{\partial x^a} \otimes dx^b$$ summed over all $a, b$.
Problem: Find $T\left(\frac{\partial}{\partial x^k}, dx^l\right)$
Wouldn't $T$ take as an input (when identified with it's corresponding multilinear map) in its first argument a one-form and in it's second argument a tangent vector?
Said more concretely since we have $$T_p(\mathbb{R}^n) \otimes T_p^*(\mathbb{R}^n) \cong L\left(T_p^*(\mathbb{R}^n) \ ,\ T_p(\mathbb{R}^n); \mathbb{R}\right)$$ we can identify $T$ with a multilinear map $$T(\omega, X_p) = \left(T^{a}_b \frac{\partial}{\partial x^a} \otimes dx^b\right)(\omega, X_p) = T^a_b \left(\frac{\partial}{\partial x^a}(\omega) \cdot {dx}^b(X_p)\right)$$
(Also note that strictly $\frac{\partial}{\partial x^a}$ doesn't take as inputs one-forms but we can identify it with an element of the bidual $T_p^{**}(\mathbb{R}^n)$ which does so we can do this without harm.
But the problem states that we need to find $T\left(\frac{\partial}{\partial x^k}, dx^l\right)$ , shouldn't it instead be to find $T\left(dx^l, \frac{\partial}{\partial x^k}\right)$?