The situation seems more complicated.

In fact, let $V$ be the vector bundle $\gamma_1 \oplus \mathbb{R}^{2k-1}$ over $S^1$.

Then $TP(V)$ is isomorphic (via the choice of a connection) to $q^*TS^1 \oplus q^*V / L$,
where $L\subset q^*V$ is the tautological bundle.

Hence the total Stiefel-Whitney class of $TP(V)$ is $q^* w(V)\cup w(L)^{-1}$ in the algebra $H^*(TP(V),\mathbb{F}_2)$.

This ~~algebra~~ is isomorphic to $\mathbb{F}_2[x,y]/(x^2,y^{2k})$, [**EDIT**: as a module over $H^*(S^1)$] since the $\mathbb{F}_2$-cohomology spectral sequence of $P(V)\to S^1$ necessarily has zero differentials on the $E_2$ page. Here $x=q^*(w_1(\gamma_1))$. Note that $\pi_1(S^1)$ acts trivially on $H^*(P^{2k-1},\mathbb{F}_2) \simeq \mathbb{F}_2[y]/(y^{2k})$.
[**EDIT** : $y\in H^1(P(V ))$ is a class that restricts to the generator of $H^1$ of any fiber. *But this doesn't characterize it* : one may add $x$ to it. Hence the algebra structure must be determined by other means. See the comments].

But $w_1(L)\in H^1(P(V),\mathbb{F}_2)\simeq Hom(\pi_1(P(V)),\mathbb{F}_2)$ is easily checked to be $x+y$ : first note that $\pi_1(V)\simeq \mathbb{Z}\times \mathbb{Z}/2$, then that $L$ is non trivial along the section of $q$ given by $P(\gamma_1)$. Hence the $x$ summand. The $y$ summand comes from restriction to a fibre. [**EDIT** : here I may precise a choice of $y$. It is Poincaré dual to the "hyperplane section" $S^1\times P(\mathbb{R}^{2k-1})$ in $P(V)$. But this doesn't determine the multiplicative structure.]

[**EDIT** : The following calculation was wrong, due to a wrong algebra structure. See the comments for calculations with the correct one, given by $x^2=0$ and $(x+y)y^{2k-1}=0$.]

total spaceof the bundle. – BS. Feb 22 '12 at 10:21