Consider the space of non zero real homogeneous degree $2$ polynomials in three variables upto scaling. This space is $\mathbb{R} \mathbb{P}^5$. The zero set of such a polynomial gives a real curve in $\mathbb{R} \mathbb{P}^2$. Let me define the space $X$ to be the space of real curves $[f]$ and a marked point $p$, such that the curve has a singularity at $p$, i.e. $$ X := \{ ([f], p) \in \mathbb{R} \mathbb{P}^5 \times \mathbb{R} \mathbb{P}^2: f(p) =0, ~~\nabla f|_p =0 \}. $$

It can be shown that $X$ is a smooth $4$ dimensional manifold.

$\textbf{Question:}$ Is $X$ orientable or non orientable?

$\textbf{Comments:}$ I have two arguments giving me contradictory answers. First of all, note that the normal bundle to $X$ in $\mathbb{R} \mathbb{P}^5 \times \mathbb{R} \mathbb{P}^2$ is given by $$ N_X:= \gamma_{5}^* \otimes \gamma_2^{* 2} \oplus \gamma_5^* \otimes T^*\mathbb{R P}^2 \otimes \gamma_2^* \big{|}_X, $$ where $\gamma_n$ is the tautological line bundle over $\mathbb{RP}^n$. (I can justify this if someone is not convinced).

Note that $$ TX \approx \frac{T (\mathbb{R} \mathbb{P}^5 \times \mathbb{R} \mathbb{P}^2)}{N_X}. $$ It is now easy to see that the first stiefel whitney class of the tangent bundle of $X$ is zero, i.e. $w_1(TX) =0$. Hence $X$ is orientable.

On the other hand it seems to me that $~\pi: X \rightarrow \mathbb{RP}^2$ is a fibre bundle, with fibres $\mathbb{RP}^2$. One can show (using spectral sequences) that any $\mathbb{RP}^2$ bundle over $\mathbb{RP}^2$ is non orientable.

$\textbf{Proof of why $X$ is a manifold:} $ Let us take $\mathbb{R}^6$ to be the space of real polynomials in two variables of degree at most $2$. Hence we can write such an element $\rho \in \mathbb{R}^6$ as $$ \rho(x,y) = \rho_{00} + \rho_{10} x + \rho_{01} y + \frac{\rho_{20}}{2} x^2 + \rho_{11} x y + \frac{\rho_{02}}{2} y^2. $$

Now consider the map $$ \psi : \mathbb{R}^6 \times \mathbb{R}^2 \rightarrow \mathbb{R}^3 $$ given by $$ \psi(\rho; x,y) := \rho(x,y), ~\rho_{x}(x,y), ~\rho_y(x,y). $$ It is easy to see that at $(\rho,0,0) $ the map $\psi$ is transverse to the zero set (in the usual calculus sense). Hence $\psi$ is transverse to the zero set (it is easy to see taking $(x,y) = (0,0)$ was without any loss of generality).

Its now easy to see that the space $X$ defined will also be a manifold; by writing the evaluation map and the vertical derivative in a coordinate chart, it reduces to the above calculation.