For any smooth fiber bundle

$$ F\hookrightarrow P \stackrel{\pi}{\to} M $$

we have a short exact sequence of vector bundles over $P$

$$ 0\to VTP\to TP \to \pi^* TM\to 0, $$

where $VTP$ denotes the vertical tangent bundle defined as the kernel of the differential of $\pi$. If the bundle is holomorphic then the above is a short exact sequence of complex vector bundles and we deduce

$$ c(TP)= c(VTP)\cdot \pi^* c(TM). $$

The classical Euler exact sequence argument shows that when $P=\mathbb{P}(V)$ that $\newcommand{\bC}{\mathbb{C}}$

$$ \gamma^*\otimes \pi^*V \cong \underline{\bC}\oplus VTP, $$

where $\underline{\bC}$ denotes the trivial line bundle. Hence

$$ c(TP)= c(\gamma^*\otimes \pi^*V)\cdot \pi^* c(TP). $$

Note. The original answer had an error that I have now corrected. (Hat tip to abx).

For any smooth fiber bundle

$$ F\hookrightarrow P \stackrel{\pi}{\to} M $$

we have a short exact sequence of vector bundles over $P$

$$ 0\to VTP\to TP \to \pi^* TM\to 0, $$

where $VTP$ denotes the vertical tangent bundle defined as the kernel of the differential of $\pi$. If the bundle is holomorphic then the above is a short exact sequence of complex vector bundles and we deduce

$$ c(TP)= c(VTP)\cdot \pi^* c(TM). $$

The classical Euler exact sequence argument shows that when $P=\mathbb{P}(V)$ that $\newcommand{\bC}{\mathbb{C}}$

$$ \gamma^*\otimes \pi^*V \cong \underline{\bC}\oplus VTP, $$

where $\underline{\bC}$ denotes the trivial line bundle. Hence

$$ c(TP)= c(\gamma^*\otimes \pi^*V)\cdot \pi^* c(TP). $$

In Section I.3 of Fulton-Lang Riemann-Roch algebra you can find an explicit formula for $c_k(L\otimes E)$, $L$ line bundle and $E$ vector bundle of rank $m$. More precisely

$$ c_k(L\otimes E)=\sum_{j=1}^k \binom{m-j}{k-j} c_j(E)c_1(L)^{k-j}. $$ Note. The original answer had an error that I have now corrected. (Hat tip to abx).

Your guess is correct. (The exponent should be $k+1$, even when the base is a point.) For any smooth fiber bundle

$$ F\hookrightarrow P \stackrel{\pi}{\to} M $$

we have a short exact sequence of vector bundles over $P$

$$ 0\to VTP\to TP \to \pi^* TM\to 0, $$

where $VTP$ denotes the vertical tangent bundle defined as the kernel of the differential of $\pi$. If the bundle is holomorphic then the above is a short exact sequence of complex vector bundles and we deduce

$$ c(TP)= c(VTP)\cdot \pi^* c(TM). $$

This leads to your guess since the classical Euler exact sequence argument shows that

$$ c\bigl(\; V T\mathbb{P}(V)\;\bigr)= \bigl(\; 1+c_1(\gamma^*)\;\bigr)^{k+1}. $$

(Note the exponent is $k+1$.)

For any smooth fiber bundle

$$ F\hookrightarrow P \stackrel{\pi}{\to} M $$

we have a short exact sequence of vector bundles over $P$

$$ 0\to VTP\to TP \to \pi^* TM\to 0, $$

where $VTP$ denotes the vertical tangent bundle defined as the kernel of the differential of $\pi$. If the bundle is holomorphic then the above is a short exact sequence of complex vector bundles and we deduce

$$ c(TP)= c(VTP)\cdot \pi^* c(TM). $$

The classical Euler exact sequence argument shows that when $P=\mathbb{P}(V)$ that $\newcommand{\bC}{\mathbb{C}}$

$$ \gamma^*\otimes \pi^*V \cong \underline{\bC}\oplus VTP, $$

where $\underline{\bC}$ denotes the trivial line bundle. Hence

$$ c(TP)= c(\gamma^*\otimes \pi^*V)\cdot \pi^* c(TP). $$

Note. The original answer had an error that I have now corrected. (Hat tip to abx).