Let $\pi : PE \to B$ be a projectivised rank $n$ vector bundle. In this MathOverflow answer, Michael Thaddeus says that for such a bundle, $T_{\pi} = \operatorname{Hom}(O(-1), E/O(-1)) \cong E(1)/O$, so $\bigwedge^{n-1}T_{\pi} \cong (\bigwedge^nE)(n)$ canonically.
I tried to construct such an identification, but I wasn't able to. I try the following : let $E$ a complex vectorial space with dimension $n$ and $S$ a complex line in $E$. I try to find a canonical isomorphism between $\wedge^{n-1}E/S$ and $\wedge^{n} E \otimes S^{n}$. If $e_{1}$ is a base of $S$ completed in a base $e_{1}, ..., e_{n}$ of $E$, i construct $\phi$ linear which send $e^{2} \wedge ... \wedge e^{n} \in \wedge^{n-1}E/S$ on $e^{1} \wedge ... \wedge e^{n} \otimes (e^{1})^{\otimes n}$ (the $e^{*}$ means the dual basis of $e$). Unfortunaltely, it depends of the choice of $e$. Could you help me?
I wish you a good day.
Note that $T_{\pi} \cong E(1)/O$ while $\bigwedge^{n-1}T_{\pi} \cong (\bigwedge^nE)(n)\cong\bigwedge^nE(1)$. This explains why your vector space analysis is off: you should be trying to construct an isomorphism between $\bigwedge^{n-1}(E/S)$ and $\bigwedge^nE$ - here $E$ plays the role of $E(1)$ and $S$ plays the role of $O$. While an isomorphism exists, it won't be canonical as you'll have to choose a non-zero element of $S$. On the other hand, there is a canonical isomorphism $S\otimes\bigwedge^{n-1}(E/S) \to \bigwedge^nE$. This gives us an indication of what is happening globally.
Since $T_{\pi} \cong E(1)/O$, we have a short exact sequence $0 \to O \to E(1) \to T_{\pi} \to 0$. It follows that we have canonical isomorphisms $$\bigwedge\nolimits^{\!n}E(1) \cong \bigwedge\nolimits^{\!1}O\otimes\bigwedge\nolimits^{\!n-1}T_\pi \cong O\otimes\bigwedge\nolimits^{\!n-1}T_{\pi} \cong \bigwedge\nolimits^{\!n-1}T_{\pi}.$$
Therefore $\bigwedge^{n-1}T_\pi\cong \bigwedge^nE(1) \cong (\bigwedge^nE)(n)$.
The fact that the isomorphism between determinant bundles is canonical follows from the corresponding statement for vector bundles which is discussed in this question.