No.
$\hat p\sin(kx)$ is not a multiple of itself. It is true that
$\hat p\,e^{\pm i k x}= \pm\hbar k e^{\pm i k x}$ but a (complex) linear combination of two eigenfunctions is only an eigenfunction if both eigenfunctions in the sum have the same eigenvalue, which is NOT the case here as the eigenvalues $\pm \hbar k$ differ by a sign.
$\sin(kx)$ is an eigenfunction of $\hat p^2$ (although be careful as it is not a normalizable eigenfunction in the sense that $\int_{-\infty}^\infty \sin^2(kx) dx$ is not finite).
In response to comments:
Let $\hat T$ be your favorite operator, and let $\phi_A(x)$ and $\phi_B(x)$ be such that
$$
\hat T\phi_A(x)=A\phi_A(x)\, ,\qquad
\hat T\phi_B(x)=B\phi_B(x)\, .
$$
Take the complex linear combination
$$
\psi(x)=\alpha \phi_A(x)+ \beta \phi_B(x)
$$
and look at
\begin{align}
\hat T\psi(x)&= A\alpha \psi_A(x)+B\beta\phi_B(x)\\
&=A(\alpha \psi_A(x)+\beta\phi_B(x))+(B-A)\beta \phi_B(x)\\
&=A\psi(x)+(B-A)\beta\phi_B(x)
\end{align}
For this to be a multiple of the original $\psi(x)$ one must "cancel" the extra $\phi_B(x)$ term and so have $A=B$, i.e. the eigenvalues are the same, or $\beta=0$, meaning you did not have a linear combination to start with.
The calculation of average values is done using
\begin{equation}
\int dx\, \psi^*(x) \left[ \hat T\psi(x)\right]\,
\end{equation}
You can use the explicit expression of $\psi(x)$ above and find,
if your functions $\phi_A(x)$ and $\phi_B(x)$ are orthonormal in the sense
that
$$
\int dx \phi_i^*(x) \phi_j(x)=\delta_{ij}
$$
that the average value of $\hat T$ comes out as
$\alpha^*\alpha A+\beta^*\beta B$. Of course if $A=B$ and $\alpha^*\alpha +\beta^*\beta=1$ then the average value is just the eigenvalue, but this is really a special case of the general expression for the average value. The average value makes sense as the integral above for any $\psi(x)$, irrespective of whether or not one know the expansion of $\psi(x)$ in terms of eigenfunctions of $\hat T$.
In the case of $\sin(n\pi x/L)$ and $\hat p$, the mathematical machinery can bypassed by a physical argument. The quantity
$$
\psi^*(x)\left[\hat p \psi(x)\right]= i\hbar (n\pi/L) \sin(n\pi x/L)\cos(n \pi x/L)
$$
is purely imaginary, so the integral will also be purely imaginary. However, $\hat p$ is an observable so its average values must be real. This is a contradiction unless the average value is $0$. This "trick" will apply whenever the wavefunction is real and the observable is represented by an operator of the form $i\times$(something real). Another situation where this trick would be useful would be the average value of $\hat p$ for a wave function that is a real linear combination of harmonic oscillator states.
