= Solution
Pullback and <tensor product> preserve <holomorphic line bundles> and their isomorphisms. Interpret negative tensor powers of $L$ as powers of $L^*$. Thus the formula defines a group homomorphism on isomorphism classes.
To prove injectivity, suppose $p^*M\otimes L^{\otimes n}$ is trivial. Restrict it to any fibre. A line pulled back from its base point is trivial there, so its fibre restriction is $\mathcal O_{\mathbb P^1}(n)$. By the <Picard group> classification, $n=0$. It remains to show that $p^*M$ trivial implies $M$ trivial.
Let $\sigma$ be a nowhere-zero holomorphic section of $p^*M$. On a sufficiently small open set where both $E$ and $M$ are trivial, write $\sigma=a_i(x,[v])p^*e_i$ for a frame $e_i$ of $M$. For each fixed $x$, $a_i(x,\cdot)$ is a <holomorphic function> on the compact <complex projective line>, hence is constant. It is therefore $b_i(x)$; evaluating at a constant projective point in the local trivialization shows $b_i$ is holomorphic in $x$. The section is nowhere zero, so each $b_i$ is nowhere zero. Its overlap laws are exactly those of a section of $M$, and $b_ie_i$ glue to a global holomorphic frame. Thus $M$ is trivial.
The kernel consists only of $(\mathcal O_X,0)$, proving
$$
\boxed{\operatorname{Pic}_{\rm hol}(X)\times\mathbb Z
\hookrightarrow\operatorname{Pic}_{\rm hol}(\mathbb P(E)).}
$$
This is the <Picard injection for holomorphic projective bundles>. The argument uses local fibrewise constancy, so no global section of the <projective bundle> is required.
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