The answer is negative for all pairs $(k,d)$ with $k+1\leq d\leq 2k$, as long as $k \ge 2$.

Brehm [1] constructed a triangulation of the Möbius strip that does not admit a *geometric* (simplexwise linear) embedding into $\mathbb{R}^3$.

More generally, for every pair $(k,d)$ with $k+1\leq d\leq 2k$, Brehm and Sarkaria [2] constructed an example of a $k$-dimensional simplicial complex that admits a piecewise
linear embedding into $\mathbb{R}^d$, but no geometric embedding.

Moreover, for any given integr $r \geq 0$, there is such a $K$ such that even the $r$-fold barycentric subdivision of $K$ is not geometrically embeddable into $R^d$.

Furthermore,
for certain values of the parameters, e.g., for $k=d-1$ and $d\geq 5$, it is known that
there is no recursive bound on the complexity of the subdivision needed to embed a finite
$k$-dimensional simplicial complex piecewise linearly into $\mathbb{R}^d$ (see [3, Corollary 1.2]).

[1] U. Brehm. *A nonpolyhedral triangulated Möbius strip*. Proc. Amer. Math. Soc., 89(3), 519–522, 1983.

[2] U. Brehm and K. Sarkaria. *Linear vs. piecewise-linear embeddability of simplicial complexes*. Tech. Report 92/52, Max-Planck-Institut für Mathematik, Bonn, 1992. Available for
download from Karanbir Sarkaria's website http://kssarkaria.org/List.htm

[3] J. Matousek, M. Tancer, and U. Wagner. Hardness of embedding simplicial complexes in $\mathbb{R}^d$. J. Eur. Math. Soc. 13, 259–295, 2011.