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The base change of your quaternion algebra will be ramified at $\mathfrak{P}$ if and only the degree of the extension of completions $K_{\mathfrak{P}}/F_{\mathfrak{p}}$ has odd degree, i.e. case 3. A general way to see this kind of property is that the invariant in $\mathbb{Q}/\mathbb{Z}$ of the class in the Brauer group gets multiplied by the degree when you base-change the corresponding algebra. A more down-to-earth proof in the case of quaternion algebra is to use the standard presentation of the division quaternion algebra over the local field $F_{\mathfrak{p}}$ and to see that it contains all the quadratic extensions of $F_{\mathfrak{p}}$; in case 3 the local extension is trivial and so locally there is no base-change at all.