What is condition for a convex polyhedron to be uniform? A uniform polyhedron has all its vertices exactly lying on a spherical surface with a certain radius. 
Condition: A convex polyhedron will be uniform (i.e. all the vertices will exactly lie on a spherical surface) if and only if its each polygonal (flat) face is a cyclic polygon (except triangle). 
I have seen this condition is satisfied by all 5 platonic solids & all 13 Archimedean solids. 
Does this condition hold good for each convex type uniform polyhedron?   
 A: I've never heard of uniformity phrased in that context. The typical definition is that the symmetry group of the polyhedron acts transitively on the vertices of the polyhedron: given any pair of vertices, there's a symmetry that takes one to the other. What you're calling "uniform" is actually a weaker condition than the standard definition of uniform.
Interestingly, it seems that uniformity will force all vertices to lie on a sphere: a symmetry of a polyhedron is an isometry of $\Bbb R^3$, a transformation that preserves lengths. As such, any isometry must preserve the length from the center of a polytope to a vertex. Since any pair of vertices can be transformed to any other, these lengths must all be equal, hence the polyhedron must have its vertices on a sphere.
The converse isn't true: any four points on a sphere determine a tetrahedron with cyclic faces, but it doesn't need to be a uniform tetrahedron.
Edit:
Per your comment about cyclic (=all vertices lying on a circle) faces, we can find a polyhedron will all cyclic quadrilateral faces, but whose vertices don't lie on a sphere.
The idea relies on the fact that an isosceles trapezoid (take an isosceles triangle, and cut off the corner on the axis of symmetry) is always cyclic. So, if we take a right square pyramid and cut off the vertex at which the four triangles meet, we have a polyhedron with six quadrilateral faces, two square, and four isosceles trapezoids. If made the pyramidal frustum especially tall, and glue another pyramidal frustum to the bottom (make them share the same square base), the resulting convex polyhedron will have cyclic faces, but need not have a sphere on which all of its vertices lie.
So, all vertices lying on a sphere implies cyclic faces, but the converse need not be true.
A: The if part of your condition is false in the general case. Any triangle is a cyclical polygon, but you can quite easily find a convex polyhedron having all triangular faces, but whose vertices cannot lie on a spherical surface. For instance, the octahedron whose vertices in $\mathbb{R}^3$ are:
$$A=(1,0,0)\\B=(0,1,0)\\C=(-1,0,0)\\D=(0,-1,0)\\T=(0,0,1)\\P=(0,0,-2)$$
(Its faces are $\stackrel{\small\triangle}{ABT},\stackrel{\small\triangle}{BCT},\stackrel{\small\triangle}{CDT},\stackrel{\small\triangle}{DAT},\stackrel{\small\triangle}{ABP},\stackrel{\small\triangle}{BCP},\stackrel{\small\triangle}{CDP},\stackrel{\small\triangle}{DAP}$ )
There's only one spherical surface $\mathcal{S}$ passing for $A,B,C,D,T$, which is $x^2+y^2+z^2=1$, i.e. center in $O=(0,0,0)$ and radius $r=1$.
But $P\notin\mathcal{S}$.
Added
It holds true, on the other hand, that a uniform convex polyhedron (according to this definition) always has cyclic faces (the intersection of a spherical surface with an affine plane is either a circumference or a point)
