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 $k+1$.)