When the Shortcut Does Not Fit

Nothing special — the skip carries $x$ across untouched, and the addition lines up the $128$ channels the two sides have in common, leaving the remaining $128$ exactly as $F(x)$ produced them.

Concatenate $x$ onto $F(x)$ along the depth axis instead of adding them, the way an Inception block joins its parallel paths.

Drop the skip connection in this block and use a plain convolutional block, since an identity shortcut has no meaning once the shape changes.

Send $x$ through its own $1 \times 1$ convolution with $256$ filters at stride $2$, so the shortcut arrives as $14 \times 14 \times 256$ and the addition is defined.