Orientation reversal negates Pontryagin numbers

For each partition $J$ of $k$, [F2] gives $p_J(T(-M))=p_J(TM)$ as cohomology classes: reversing the orientation changes only the ray datum of [F1] and leaves the underlying smooth manifold and its tangent bundle unchanged. Consequently, using the definition of the Pontryagin number and the sign rule of [F3], $p_J[-M]=⟨p_J(T(-M)),[-M]⟩=⟨p_J(TM),-[M]⟩=-⟨p_J(TM),[M]⟩=-p_J[M],$ where the middle equality is the linearity of the Kronecker pairing in its second variable [F4].

The Stiefel-Whitney numbers are unchanged: the canonical mod-two fundamental class used in Stiefel-Whitney numbers of a closed manifold depends only on the smooth structure, and the tangent Stiefel-Whitney classes are computed from $TM$ alone, so replacing $o$ by $-o$ alters neither the classes nor the fundamental class [F2, F5]; equivalently, over ${\mathbb{F}}_2$ the orientation sign $-1$ equals $1$.

Specialize to $M={\mathbb{C}\mathbb{P}}^2$ with its complex orientation. By the A-page tangent-bundle lemma [F5], $p_1[{\mathbb{C}\mathbb{P}}^2]=3$ and it is the only Pontryagin number in degree four, the partition being $(1)$; step 1.1 gives $p_1[-{\mathbb{C}\mathbb{P}}^2]=-3$. Both values are nonzero, so neither orientation is an oriented boundary by [F5], and the two orientations are distinguished by $p_1$ even though the underlying unoriented manifold and all its Stiefel-Whitney numbers are the same.

Boundary cases. For $k=0$ the manifold is a finite set of signed points, the only Pontryagin number is $p_{\mathrm{\varnothing }}=⟨1,[M]⟩$, the signed count, and step 1.1 gives the sign reversal of that count; the empty manifold has value $0$. Higher Pontryagin classes of ${\mathbb{C}\mathbb{P}}^2$ vanish because $p(T{\mathbb{C}\mathbb{P}}^2)=1+3x^2$ with $x^3=0$ in the truncated ring, so there is no second number to test; for a general $M$ all partitions $J$ of $k$ are covered by step 1.1. No choice beyond the cited suppliers is used.